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Which of the following queries can be used to get multiple column values from a table?
Using the SELECT Statement to Retrieve Data in SQL To retrieve multiple columns from a table, you use the same SELECT statement. The only difference is that you must specify multiple column names after the SELECT keyword, and separate each column by a comma.
What can a subquery return?
A subquery selects and returns values to the first or outer SELECT statement. A subquery can return no value, a single value, or a set of values, as follows: If a subquery returns no value, the query does not return any rows. Such a subquery is equivalent to a null value.
How many values can a subquery return?
Subquery returned more than 1 value. This is not permitted when the subquery follows =, != , <, <= , >, >= or when the subquery is used as an expression.
How to select multiple columns from a subquery?
SELECT a.salesorderid, a.orderdate, s.orderdate, s.salesorderid FROM A a OUTER APPLY (SELECT top (1) * FROM B b WHERE a.salesorderid = b.salesorderid) as s WHERE A.Date BETWEEN ‘2000-1-4’ AND ‘2010-1-4’ This will select exactly one record from B for each SalesOrderId.
How to use JSON _ query in SQL Server?
For more info about the optional path mode specification (lax or strict), see JSON Path Expressions (SQL Server). Returns the entire JSON text. Returns the entire JSON text. Not an object or array. Use JSON_VALUE instead. Not an object or array. Use JSON_VALUE instead. Not an array. Property does not exist. JSON_QUERY returns a valid JSON fragment.
How to return result set as JSON array?
A couple of examples: Also if you want selected field from table and aggregated then as array . The result will come . Thanks for contributing an answer to Stack Overflow!
Can a JSON query return a valid JSON fragment?
JSON_QUERY returns a valid JSON fragment. As a result, FOR JSON doesn’t escape special characters in the JSON_QUERY return value. If you’re returning results with FOR JSON, and you’re including data that’s already in JSON format (in a column or as the result of an expression), wrap the JSON data with JSON_QUERY without the path parameter.