prepare() not working

3 Views

When I do this: 

$transients = $wpdb->get_col(
    "
    SELECT     option_name
    FROM       $wpdb->options
    WHERE      option_name
    LIKE       '_transient_wb_tt_%'
    "
);

It works fine, but when I use prepare like so:

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$transients = $wpdb->get_col( $wpdb->prepare(
    "
    SELECT     option_name
    FROM       %s
    WHERE      option_name
    LIKE       '_transient_wb_tt_%'
    ",
    $wpdb->options
) );

It doesn’t work, what am I doing wrong here?

Post Views: 3

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2 comments

  1. I agree with @bainternet. You don’t need $wpdb->prepare. There isn’t any user supplied content.

    The answer to the question is that to get a wildcard % to pass through prepare you need to double it in your code. 

    LIKE  '_transient_wb_tt_%%'

    Try that or this if you want a good look at the generated query:

    var_dump($wpdb->prepare("
    SELECT     option_name
    FROM       %s
    WHERE      option_name
    LIKE       '_transient_wb_tt_%%'
    ",
    'abc')); 
die;

    Other than being unnecessary, using $wpdb->prepare like this won’t work. The attempt to use prepare to swap in the tablename will result in a tablename with quotes around it. That is invalid SQL. The query should be simple:

    SELECT     option_name
FROM       {$wpdb->options}
WHERE      option_name
LIKE       '_transient_wb_tt_%%'

  2. From Codex

    $wild = '%';
$find = 'only 43% of planets';
$like = $wild . $wpdb->esc_like( $find ) . $wild;
$sql  = $wpdb->prepare( "SELECT * FROM $wpdb->posts WHERE post_content LIKE %s", $like );

    The essential part is -> “$like = $wild . $wpdb->esc_like( $find ) . $wild;”