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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

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));
| improve this answer | |
  • 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 '15 at 23:11
  • this example had single quotes ', instead of ` wrapped around the columns – Andre Baumeier Mar 25 '15 at 9:53

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