How to Perform Left Outer Joins in LINQ Using Extension Methods?

Use the linq expansion method to execute the left external connection

Left outer connection is a connection operation. It returns all the lines in the left table (called the "Father" table), and the right table (called the "Zi" table) only matched row. If you can't find a matching line in the sub -table, the return value of the line in the parent table is null.

In linq, a method of executing the left and outer connection is to use the

method, which creates an anonymous type sequence that contains the attributes in the parent table and the sub -table. However, the more convenient way to achieve the same results is to use the

and Join expansion methods. GroupJoin SelectMany The left -outer connection syntax using the expansion method is as follows:

In this grammar:

<code class="language-csharp">var query = parentTable.GroupJoin(
    childTable, 
    parentKeySelector,
    childKeySelector,
    (parent, child) => new { Parent = parent, Child = child.DefaultIfEmpty() })
    .SelectMany(
        x => x.Child,
        (x, y) => new { Parent = x.Parent, Child = y });</code>

and

are the tables to be connected.
• parentTable and childTable are Lambda expressions specified to connect the keys for two tables.
• The anonymous type in the method combines the father and the son. parentKeySelector childKeySelector Methods to ensure that the NULL value is returned to the NULL value for a child who does not match the parent.
• The method flattes the sequence of anonymous types into a single sequence. Each element represents a line in the parent table and the corresponding sub -line (if there is no matching child, it is null). GroupJoin
• Example DefaultIfEmpty()
• Considering the left and external connection query in the following SQL: SelectMany

You can use the extension method as follows. This query is as follows:

This query will return an anonymous type sequence, each of which contains the attributes of the FOO and BAR table. If the line in the FOO table does not match in the BAR table, the corresponding BAR attribute will be NULL.

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