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I'm trying to substitute a variable in a db_query but keep on getting this error:

Recoverable fatal error: Argument 2 passed to db_query() must be of the type array, string given

It's a single item string that I'm trying to insert:

$primary_key = db_query("SELECT `COLUMN_NAME` FROM `information_schema`.`COLUMNS` WHERE (`TABLE_SCHEMA` = 'mydatabase') AND (`TABLE_NAME` = '%s') AND (`COLUMN_KEY` = 'PRI')",$mytable)->fetchAll(PDO::FETCH_ASSOC);

I've tried wrapping the variable in an array but then I get no results. How do I substitute my variable here?

1 Answer 1

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Drupal 7 doesn't use the '%' place holders. Try this:

$primary_key = db_query("SELECT `COLUMN_NAME` FROM `information_schema`.`COLUMNS` WHERE (`TABLE_SCHEMA` = 'mydatabase') AND (`TABLE_NAME` = :tbl_name) AND (`COLUMN_KEY` = 'PRI')", array(':tbl_name' => $mytable));
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  • Got a MySQL error on this one.Looks like it's changing the substitution: "('COLUMN_KEY' = 'PRI'); Array ( [:tbl_name] => store ) in"
    – Ken J
    Mar 24, 2015 at 23:11
  • this example had single quotes ', instead of ` wrapped around the columns Mar 25, 2015 at 9:53

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