Data Joins
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Data Joins |
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Joins |
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Introduction |
When studying relationships, we reviewed techniques of making data from one table available to the records of another table. This demonstrated to reduce data duplication and mistakes. Another issue that involves the combination of tables consists of creating records from more than one table and making the result into a single list. This is the basis of data joins.
A data join is a technique of creating a list of records from more than one table, using all columns from all tables involved, or selecting only the desired columns from one or all of the tables involved. This means that a data join is essentially created in three steps:
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The Tables of a Join |
Before creating a join, you must have the tables that would be involved. The tables are created using the techniques we have seen in previous lessons. It is also important to create a primary key for each table. The parent table would usually need only this primary key that would be used to "link" it to a child table. Here is an example of such a table:
If needed, you can then create the necessary records for the table. Here is an example:
When creating the child table, remember to create a column that would serve as the link to the parent table. By a (good) habit as we saw when studying relationships, the name and the data type of this column is the same as the primary key of the parent table. Here is an example of a child table that would be joined to the above parent table:
Once again, if necessary, you can add the needed records to the table. Here is an example:
using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Windows.Forms;
using System.Data.SqlClient;
namespace AltairRealtors7
{
public partial class AltairRealtors : Form
{
public AltairRealtors()
{
InitializeComponent();
}
void CreateDatabase()
{
string csAltairRealtors = "Data Source=(local);" +
"Database='AltairRealtors2';" +
"Integrated Security='SSPI';";
string strAltairRealtors = "";
SqlConnection cnnAltairRealtors = null;
SqlCommand cmdAltairRealtors = null;
using (cnnAltairRealtors =
new SqlConnection("Data Source=(local);" +
"Integrated Security='SSPI';"))
{
strAltairRealtors = "CREATE DATABASE AltairRealtors2;";
cmdAltairRealtors = new SqlCommand(strAltairRealtors,
cnnAltairRealtors);
cnnAltairRealtors.Open();
cmdAltairRealtors.ExecuteNonQuery();
MessageBox.Show("The AltairRealtors2 database has been created");
}
using (cnnAltairRealtors =
new SqlConnection(csAltairRealtors))
{
strAltairRealtors =
"CREATE TABLE PropertyTypes" +
"(" +
"PropertyTypeID int identity(1,1) NOT NULL," +
"PropertyType varchar(20), " +
"CONSTRAINT PK_PropertyTypes PRIMARY KEY (PropertyTypeID));";
cmdAltairRealtors = new SqlCommand(strAltairRealtors,
cnnAltairRealtors);
cnnAltairRealtors.Open();
cmdAltairRealtors.ExecuteNonQuery();
MessageBox.Show("The PropertyTypes table has been created");
}
using (cnnAltairRealtors = new SqlConnection(csAltairRealtors))
{
strAltairRealtors =
"INSERT INTO PropertyTypes(PropertyType) " +
"VALUES('Condominium'); " +
"INSERT INTO PropertyTypes(PropertyType) " +
"VALUES('Single Family'); " +
"INSERT INTO PropertyTypes(PropertyType) " +
"VALUES('Townhouse'); " +
"INSERT INTO PropertyTypes(PropertyType) " +
"VALUES('Unknown');";
cmdAltairRealtors = new SqlCommand(strAltairRealtors,
cnnAltairRealtors);
cnnAltairRealtors.Open();
cmdAltairRealtors.ExecuteNonQuery();
MessageBox.Show("Some records have been created in the Property Types table.");
}
using (cnnAltairRealtors = new SqlConnection(csAltairRealtors))
{
strAltairRealtors =
"CREATE TABLE Conditions " +
"( " +
"ConditionID int identity(1,1) NOT NULL, " +
"Condition varchar(20), " +
"CONSTRAINT PK_Conditions PRIMARY KEY (ConditionID));";
cmdAltairRealtors = new SqlCommand(strAltairRealtors,
cnnAltairRealtors);
cnnAltairRealtors.Open();
cmdAltairRealtors.ExecuteNonQuery();
MessageBox.Show("The Conditions table has been created");
}
using (cnnAltairRealtors = new SqlConnection(csAltairRealtors))
{
strAltairRealtors =
"INSERT INTO Conditions(Condition) " +
"VALUES('Excellent'); " +
"INSERT INTO Conditions(Condition) " +
"VALUES('Good'); " +
"INSERT INTO Conditions(Condition) " +
"VALUES('Bad Shape'); " +
"INSERT INTO Conditions(Condition) " +
"VALUES('Unknown');";
cmdAltairRealtors = new SqlCommand(strAltairRealtors,
cnnAltairRealtors);
cnnAltairRealtors.Open();
cmdAltairRealtors.ExecuteNonQuery();
MessageBox.Show("Some records have been created in the Conditions table.");
}
using (cnnAltairRealtors =
new SqlConnection("Data Source=(local);" +
"Database='AltairRealtors2';" +
"Integrated Security='SSPI';"))
{
strAltairRealtors =
"CREATE TABLE Properties" +
"(" +
"PropertyID int identity(1,1) NOT NULL," +
"PropertyNumber char(6)," +
"Address varchar(100)," +
"City varchar(50)," +
"State char(2)," +
"ZIPCode varchar(12)," +
"PropertyTypeID varchar(40)," +
"ConditionID varchar(32)," +
"Bedrooms smallint," +
"Bathrooms float," +
"FinishedBasement bit," +
"IndoorGarage bit," +
"Stories smallint," +
"YearBuilt smallint," +
"MarketValue money, " +
"CONSTRAINT PK_Properties PRIMARY KEY (PropertyID));";
cmdAltairRealtors = new SqlCommand(strAltairRealtors,
cnnAltairRealtors);
cnnAltairRealtors.Open();
cmdAltairRealtors.ExecuteNonQuery();
MessageBox.Show("The Properties table has been created");
}
using (cnnAltairRealtors = new SqlConnection(csAltairRealtors))
{
strAltairRealtors =
"INSERT INTO Properties(PropertyNumber, Address, City, State," +
"ZIPCode, PropertyTypeID, ConditionID, Bedrooms, Bathrooms," +
"FinishedBasement, IndoorGarage, Stories, YearBuilt, MarketValue)" +
"VALUES('524880', '1640 Lombardo Ave', 'Silver Spring', 'MD'," +
"'20904', 2, 2, 4, 2.5, 3, 1, 3, 1995, 495880.00);" +
"INSERT INTO Properties(PropertyNumber, Address, City, State," +
"ZIPCode, PropertyTypeID, ConditionID, Bedrooms, Bathrooms," +
"FinishedBasement, IndoorGarage, Stories, YearBuilt, MarketValue)" +
"VALUES('688364', '10315 North Hacht Rd', 'College Park', 'MD'," +
"'20747', 2, 1, 4, 3.5, 3, 1, 2, 2000, 620724.00);" +
"INSERT INTO Properties(PropertyNumber, Address, City, State," +
"ZIPCode, PropertyTypeID, ConditionID, FinishedBasement," +
"Stories, MarketValue)" +
"VALUES('611464', '6366 Lolita Drive', 'Laurel', 'MD'," +
"'20707', 2, 2, 1, 2, 422625.00);" +
"INSERT INTO Properties(Address, City, PropertyTypeID," +
"Bedrooms, MarketValue)" +
"VALUES('9002 Palasko Hwy', 'Tysons Corner'," +
"1, 2, 422895.00);" +
"INSERT INTO Properties(PropertyNumber, State," +
"ZIPCode, Bedrooms, YearBuilt, MarketValue)" +
"VALUES('420115', 'DC', '20011', 2, 1982, 312555);" +
"INSERT INTO Properties(PropertyNumber, City, ZIPCode," +
"PropertyTypeID, Bedrooms, YearBuilt, MarketValue)" +
"VALUES('917203', 'Alexandria', '22024'," +
"2, 3, 1965, 345660.00);" +
"INSERT INTO Properties(PropertyNumber, Address, City, State," +
"PropertyTypeID, ConditionID, Bedrooms, Bathrooms, MarketValue)" +
"VALUES('200417', '4140 Holisto Crt', 'Germantown', 'MD'," +
"1, 1, 2, 1, 215495.00);" +
"INSERT INTO Properties(PropertyNumber, Address, City, State," +
"ZIPCode, PropertyTypeID, ConditionID, Bedrooms, Bathrooms," +
"FinishedBasement, IndoorGarage, Stories, YearBuilt, MarketValue)" +
"VALUES('927474', '9522 Lockwood Rd', 'Chevy Chase', 'MD'," +
"'20852', 3, 3, 3, 2.5, 3, 0, 3, 1992, 415665.00);" +
"INSERT INTO Properties(PropertyNumber, Address, City, State," +
"ZIPCode, PropertyTypeID, ConditionID, Bedrooms, Bathrooms," +
"FinishedBasement, IndoorGarage, Stories, YearBuilt, MarketValue)" +
"VALUES('207850', '14250 Parkdoll Rd', 'Rockville', 'MD'," +
"'20854', 3, 2, 3, 2.5, 2, 1, 2, 1988, 325995.00);" +
"INSERT INTO Properties(City, PropertyTypeID, Bedrooms," +
"YearBuilt, MarketValue)" +
"VALUES('Washington', 3, 4, 1975, 366775.00);" +
"INSERT INTO Properties(PropertyNumber, Address, City, State," +
"ZIPCode, PropertyTypeID, ConditionID, Bedrooms, Bathrooms," +
"YearBuilt, MarketValue)" +
"VALUES('288540', '10340 Helmes Street #408', 'Silver Spring', 'MD'," +
"'20906', 1, 2, 1, 1, 2000, 242775.00);" +
"INSERT INTO Properties(PropertyNumber, Address, City, State," +
"ZIPCode, PropertyTypeID, ConditionID, Bedrooms, Bathrooms," +
"FinishedBasement, IndoorGarage, Stories, YearBuilt, MarketValue)" +
"VALUES('247472', '1008 Coppen Street', 'Silver Spring', 'MD'," +
"'20906', 2, 1,3, 3, 3, 1, 3, 1996, 625450.00);" +
"INSERT INTO Properties(City, ZIPCode, PropertyTypeID, " +
"Stories, YearBuilt, MarketValue)" +
"VALUES('Chevy Chase', '20956', 2, 3, 2001, 525450.00);" +
"INSERT INTO Properties(Address, City, State," +
"PropertyTypeID, ConditionID, Bedrooms, MarketValue)" +
"VALUES('686 Herod Ave #D04', 'Takoma Park', 'MD'," +
"1, 1, 2, 360885.00);" +
"INSERT INTO Properties(PropertyNumber, Address, City, State," +
"ZIPCode, PropertyTypeID, ConditionID, Bedrooms, Bathrooms," +
"FinishedBasement, IndoorGarage, Stories, YearBuilt, MarketValue)" +
"VALUES('297446', '14005 Sniders Blvd', 'Laurel', 'MD'," +
"'20707', 3, 4," +
"4, 1.5, 3, 1, 2, 2002, 412885.00);" +
"INSERT INTO Properties(City, ZIPCode, ConditionID, Bedrooms," +
"Stories, YearBuilt)" +
"VALUES('Silver Spring', '20905', 2," +
"4, 2, 1965);" +
"INSERT INTO Properties(PropertyNumber, Address, City, State," +
"ZIPCode, PropertyTypeID, ConditionID, Bedrooms, Bathrooms," +
"FinishedBasement, IndoorGarage, Stories, YearBuilt, MarketValue)" +
"VALUES('924792', '680 Prushia Rd', 'Washington', 'DC'," +
"'20008', 2, 2, 5, 3.5, 3, 0, 3, 2000, 555885.00);" +
"INSERT INTO Properties(PropertyNumber, Address, City, State," +
"ZIPCode, PropertyTypeID, ConditionID, Bedrooms, Bathrooms," +
"FinishedBasement, IndoorGarage, Stories, YearBuilt, MarketValue)" +
"VALUES('294796', '14688 Parrison Street', 'College Park', 'MD'," +
"'20742', 2, 1, 5, 2.5, 2, 1, 2, 1995, 485995.00);" +
"INSERT INTO Properties(PropertyNumber, Address, City, State," +
"ZIPCode, PropertyTypeID, ConditionID, Bedrooms, Bathrooms," +
"YearBuilt, MarketValue)" +
"VALUES('811155', '10340 Helmes Street #1012', 'Silver Spring'," +
"'MD', '20906', 1, 2, 1, 1, 2000, 252775.00);" +
"INSERT INTO Properties(PropertyNumber, Address, City, State," +
"ZIPCode, PropertyTypeID, ConditionID, Bedrooms, Bathrooms," +
"FinishedBasement, IndoorGarage, Stories, YearBuilt, MarketValue)" +
"VALUES('447597', '4201 Vilamar Ave', 'Hyattsville', 'MD'," +
"'20782', 3, 1, 3, 2, 2, 1, 3, 1992, 365880.00);" +
"INSERT INTO Properties(Address, ZIPCode, Bathrooms)" +
"VALUES('1622 Rombard Str', 20904, 2.5);" +
"INSERT INTO Properties(PropertyNumber, Address, City, State," +
"ZIPCode, PropertyTypeID, ConditionID, Bedrooms, Bathrooms," +
"FinishedBasement, IndoorGarage, Stories, YearBuilt, MarketValue)" +
"VALUES('297415', '980 Phorwick Street', 'Washington', 'DC'," +
"'20004', 2, 2, 4, 3.5, 3, 3, 1, 2004, 735475.00);" +
"INSERT INTO Properties(PropertyNumber, Address, City, State," +
"ZIPCode, PropertyTypeID, ConditionID, Bedrooms, Bathrooms," +
"FinishedBasement, IndoorGarage, Stories, YearBuilt, MarketValue)" +
"VALUES('475974', '9015 Marvin Crow Ave', 'Gaithersburg', 'MD'," +
"'20872', 2, 4, 4, 2.5, 3, 1, 1, 1965, 615775.00);";
cmdAltairRealtors =
new SqlCommand(strAltairRealtors,
cnnAltairRealtors);
cnnAltairRealtors.Open();
cmdAltairRealtors.ExecuteNonQuery();
MessageBox.Show("Some records have been created in the Properties table.");
}
}
private void AltairRealtors_Load(object sender, EventArgs e)
{
CreateDatabase();
}
}
}
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| Object | Name |
| propertiesBindingSource | bsProperties |
| propertiesTableAdapter | taProperties |
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private void btnClose_Click(object sender, EventArgs e)
{
Close();
}
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Visually Creating a Join |
Equipped with the necessary tables and their columns, you can create the join. To do this in the Microsoft SQL Server Management Studio, you can right-click one of the tables involved and click Open Table. This would display the Table window. You should then display the Diagram and the SQL sections. Because the foundation of a join lies on at least two tables, you should add one. To do this
Any of these actions would display the Add Table dialog box. To select a table:
Alternatively, instead of using Add Table, you can drag the child table from the Object Explorer and drop it in the Diagram section.
Here is an example of two tables that have been added:
Remember that you can drag the title bars of the tables to move them and position them to your liking. After selecting the table(s), on the Add Table dialog box, you can click Close.
If a relationship was already established between the tables, a joining line would show it.
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Programmatically Creating a Join |
In SQL code, the basic formula to create a join is:
SELECT WhatColumn(s) FROM ChildTable TypeOfJoin ParentTable ON Condition
The ChildTable factor specifies the table that holds the records that will be retrieved. It can be represented as follows:
SELECT WhatColumn(s) FROM Persons TypeOfJoin ParentTable ON Condition
The ParentTable factor specifies the table that holds the column with the primary key that will control what records, related to the child table, that will display. This factor would be represented as follows:
SELECT WhatColumn(s) FROM Persons TypeOfJoin Sexes ON Persons.SexID = Sexes.SexID
The Condition factor is a logical expression used to validate the records that will be isolated. To create the condition, you should assign the primary key column of the parent table to the foreign key column of the child table. Because both columns likely have the same name, to distinguish them, their names should be qualified. This would be done as follows:
SELECT WhatColumn(s) FROM Persons TypeOfJoin Sexes ON Persons.SexID = Sexes.SexID
The WhatColumn(s) factor of our formula allows you to make a list of the columns you want to include in your statement. As you should be aware, you can include all columns by using the * operator. Here is an example:
Practical Learning: Introducing JoinsSELECT *
FROM Persons
TypeOfJoin Sexes
ON Persons.SexID = Sexes.SexID
In this case, all columns from all tables would be included in the result. Instead of all columns, you may want a restricted list. In this case, create the list after the SELECT keyword separating them with commas. You can use the name of a column normally if that name is not duplicated in more than one column. Here is an example:
SELECT LastName, FirstName, Sex FROM Persons TypeOfJoin Sexes ON Persons.SexID = Sexes.SexID
If the same name of a column is found in more than one table, as is the case for a primary-foreign key combination, you should qualify the name of the column by preceding it with the name of its parent table followed by a period. Here are examples:
SELECT LastName, FirstName, Persons.SexID,
Sexes.SexID, Sex
FROM Persons
TypeOfJoin Sexes
ON Persons.SexID = Sexes.SexID
In fact, to make your code easier to read, you should qualify the name of each column of your SELECT statement. Here are examples:
SELECT Persons.LastName, Persons.FirstName, Persons.SexID,
Sexes.SexID, Sexes.Sex
FROM Persons
TypeOfJoin Sexes
ON Persons.SexID = Sexes.SexID
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Cross and Inner Joins |
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Introduction |
When studying data relationships, we saw the role of the primary and foreign keys in maintaining the exchange of information between two tables. This technique of linking tables plays a major part when creating a join. It allows you to decide whether you want to include all records or only isolate some of them. To respect the direction of a relationship between two tables as it is applied to a query, Transact-SQL supports three types of joins.
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Cross Joins |
A cross join creates a list of all records from both tables as follows: the first record from the parent table is associated to each record from the child table, then the second record from the parent table is associated to each record from the child table, and so on. In this case also, there is no need of a common column between both tables. In other words, you will not use the ON clause.
To create a cross join, you can replace the TypeOfJoin factor of our formula with CROSS JOIN or CROSS OUTER JOIN. Here is an example:
SELECT Persons.PersonID, Persons.FirstName, Persons.LastName,
Sexes.SexID, Sexes.Sex
FROM Persons
CROSS JOIN Sexes
GO
If you are working visually on a table, by default, after you have just added a table to another one (if no relationship was already established between both tables), the query would be automatically made a cross join. All you have to do is to select the needed columns. After selecting the columns, you can execute the query to see the result:
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Inner Joins |
Imagine you have two tables that can be linked through one's primary key and another's foreign key.
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Notice that some records in the Persons table do not have an entry for the SexID column and were marked with NULL by the database engine. When creating a query of records of the Persons table, if you want your list to include only records that have an entry, you can create it as inner join.
By default, when creating a new query, if a relationship was already established between both tables, the query is made an inner join. If there was no relationship explicitly established between both tables, you would have to edit the SQL statement. Consider the following:
Notice that, because no relationship was previously established between both tables, the join is crossed.
To create an inner join, you have two options. You can drag the primary key from the parent table and drop it on the foreign key in the child table. Here is an example:
Alternatively, you can edit the SQL statement manually to make it an inner join. To do this, you would specify the TypeOfJoin factor of our formula with the expression INNER JOIN. Here is an example:
SELECT Persons.PersonID, Persons.FirstName, Persons.LastName, Persons.SexID,
Sexes.SexID AS [Sex ID], Sexes.Sex
FROM Persons INNER JOIN Sexes ON Persons.SexID = Sexes.SexID
After creating the join, in the Diagram section, a line would be created to join the tables. You can then execute the query to see the result. This would produce:
We mentioned earlier that you could include all columns in the query. In our result, since we are more interested in the Sex of each Persons record, we would not need the SexID column from the Sexes table. Here is an example:
As mentioned earlier, notice that the result includes only records that have an entry (a non-NULL entry) in the SexID foreign key column of the Persons table.
An alternative to the INNER JOIN expression is to simply type JOIN. Here is an example:
SELECT Persons.PersonID, Persons.FirstName, Persons.LastName,
Sexes.Sex
FROM Persons
JOIN Sexes
ON Persons.SexID = Sexes.SexID
To destroy a join between two tables, if you are working in the Table window, you can right-click the line that joins the tables and click Remove. In SQL, you must modify the expressions that make up the join (the JOIN and the ON expressions).
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Outer Joins |
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Introduction |
Instead of showing only records that have entries in the child table, you may want your query to include all records, including those that are null. To get this result, you would create an outer join. You have three options.
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Left Outer Joins |
A left outer join produces all records of the child table, also called the right table. The records of the child table that do not have an entry in the foreign key column are marked as NULL.
To create a left outer join, if you are working in the Table window, in the Diagram section, right-click the line that joins the tables and click the option that would select all records from the child table (in this case, that would be Select All Rows From Persons):
Alternatively, you can replace the TypeOfJoin factor of our formula with either LEFT JOIN or LEFT OUTER JOIN. Here is an example:
SELECT Persons.PersonID, Persons.FirstName, Persons.LastName,
Sexes.SexID, Sexes.Sex
FROM Persons
LEFT OUTER JOIN Sexes
ON Persons.SexID = Sexes.SexID
GO
In both cases, the button in the middle of the line would be added an arrow that points to the parent table. You can then execute the query to see the result. Here is an example:
Notice that the result includes all records of the Persons (also called the right) table and the records that don't have an entry in the SexID column of the Persons (the right) table are marked with NULL.
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Right Outer Joins |
A right outer join considers all records from the parent table and finds a matching record in the child table. To do this, it starts with the first record of the parent table (in this case the Sexes table) and shows each record of the child table (in this case the Persons table) that has a corresponding entry. This means that, in our example, a right outer join would first create a list of the Persons records that have a 1 (Female) value for the SexID column. After the first record, the right outer join moves to the second record, and so on, each time listing the records of the child table that have a corresponding entry for the primary key of the parent table.
To visually create a right outer join in the Table window, after establishing a join between both tables, if you had previously created a left outer join, you should remove it by right-clicking the line between the tables and selecting the second option under Remove. Then, you can right-click the line that joins them and click the option that would select all records from the parent table. In our example, you would click Select All Rows From Sexes.
To create a right outer join in SQL, you can replace the TypeOfJoin factor of our formula with RIGHT JOIN or RIGHT OUTER JOIN. Here is an example:
SELECT Persons.PersonID, Persons.FirstName, Persons.LastName,
Sexes.SexID, Sexes.Sex
FROM Persons
RIGHT OUTER JOIN Sexes
ON Persons.SexID = Sexes.SexID
GO
In both cases, the button on the joining line between the tables would have an arrow that points to the child table. You can then run the query. Here is an example:
Notice that the query result starts with the first record of the parent table, also called the left table (in this case the Sexes table), and lists the records of the child table, also called the right table (in this case the Persons table), that have the entry corresponding to that first record. Then it moves to the next SexID value. Also, notice that there are no NULL records in the Sex.
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Full Outer Joins |
A full outer join produces all records from both the parent and the child tables. If a record from one table does not have a value in the other value, the value of that record is marked as NULL.
To visually create a full outer join, in the Table window, right-click the line between the tables and select each option under Remove so that both would be checked. To create a full outer join in SQL, replace the TypeOfJoin factor of our formula with FULL JOIN or FULL OUTER JOIN. Here is an example:
SELECT Persons.PersonID, Persons.FirstName, Persons.LastName,
Sexes.SexID, Sexes.Sex
FROM Persons
FULL OUTER JOIN Sexes
ON Persons.SexID = Sexes.SexID
GO
The button on the line between the tables would now appear as a square. You can then execute the query. Here is an example:
Just as we have involved only two tables in our joins so far, you can create a join that includes many tables.
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Joins and Data Analysis |
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Introduction |
As demonstrated so far and in previous lessons, the main reason for creating queries is to isolate records. This is done using conditions and criteria. Joins enhance this capability because they allow you to consider records from different tables and include them in a common SQL statement.
In the joins we have created so far, we considered all records and let the database engine list them using only the rules of joins built-in the SQL. To make such a list more useful or restrictive, you can pose your own conditions that should be respected to isolate records like a funnel. As done in previous lessons, to include a criterion in a SELECT statement, you can create a WHERE clause.
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Using Criteria |
To create a criterion in a query you create from the table view in the Microsoft SQL Server Management Studio or Microsoft Visual Studio, first select a column to display it in the Grid section. Just as reviewed in the previous lessons when creating a query, to specify a criterion, in the Criteria box corresponding to the column, type the condition using any of the operators we reviewed in previous lessons. Here is an example:
SELECT Persons.PersonID, Persons.FirstName, Persons.LastName,
Sexes.SexID, Sexes.Sex
FROM Persons LEFT OUTER JOIN
Sexes ON Persons.SexID = Sexes.SexID
WHERE Sexes.Sex = 'female'
This would produce:
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private void cbxShow_SelectedIndexChanged(object sender, EventArgs e)
{
if (cbxShow.SelectedIndex == 1)
bsProperties.Filter = "PropertyType = 'Townhouse'";
else if (cbxShow.SelectedIndex == 2)
bsProperties.Filter = "PropertyType = 'Condominium'";
else if (cbxShow.SelectedIndex == 3)
bsProperties.Filter = "PropertyType = 'Single Family'";
else
bsProperties.Filter = "";
}
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