Question

这里我有这个查询，如果我将％f和％d的值替换为1和1，它将在我的phpmyadmin面板中工作但我通过wpdb get_results运行此查询并准备它不起作用。所有值都正确传递但仍然是输出null。

$f = $wc * (($ul + 100)/100);

$capabilities = $wpdb->prefix."capabilities";

$sql = "
SELECT u.*, up.*, up.pid, ( up.ros - up.soh ) / ( %f ) as qty_suggested
FROM $wpdb->users u
INNER JOIN wp_deals_users_products up
ON u.ID = up.user_id
INNER JOIN $wpdb->usermeta um
ON u.ID = um.user_id            
WHERE up.pid = %d
AND  um.meta_key = '$capabilities' 
AND um.meta_value LIKE '%subscriber%'
";

//echo $sql; This echoes the sql which I tested in phpmyadmin sql query runner

$results = $wpdb->get_results(
    $wpdb->prepare( 
        $sql,
        $f,
        $pid
        )
    );

echo json_encode($results);

如果我直接删除prepare并运行查询，或者使用硬编码1作为％d和％f的值，我仍然会得到null。相同的查询在phpmyadmin sql部分中可以正常工作

编辑：

这是wpdb prepare function中的最终查询：

    SELECT u.*, up.*, up.pid, ( up.ros - up.soh ) / ( %f ) as qty_suggested
        FROM wp_users u
        INNER JOIN wp_deals_users_products up
        ON u.ID = up.user_id
        INNER JOIN wp_usermeta um
        ON u.ID = um.user_id            
        WHERE up.pid = %d
        AND  um.meta_key = 'wp_capabilities' 
        AND um.meta_value LIKE '%subscriber%'

Answer 1

您在LIKE附近使用了错误的语法。

你必须这样使用。

$like_str = 'subscriber';
$f = $wc * (($ul + 100)/100);

$capabilities = $wpdb->prefix."capabilities";

    $sql = "
    SELECT u.*, up.*, up.pid, ( up.ros - up.soh ) / ( %f ) as qty_suggested
    FROM $wpdb->users u
    INNER JOIN wp_deals_users_products up
    ON u.ID = up.user_id
    INNER JOIN $wpdb->usermeta um
    ON u.ID = um.user_id            
    WHERE up.pid = %d
    AND  um.meta_key = '$capabilities' 
    AND um.meta_value LIKE %s";




    $query = $wpdb->prepare( $sql,$f,$pid,'%' . $wpdb->esc_like($like_str) . '%');



    $result = $wpdb->get_results($query);
    echo json_encode($result);

Answer 2

看起来valirables $f和$pid的值未正确显示。打印出最终查询以查看变量是否正确传递给＆＃39; $ sql＆＃39;像这样：

$sql = "
SELECT u.*, up.*, up.pid, ( up.ros - up.soh ) / ( %f ) as qty_suggested
FROM $wpdb->users u
INNER JOIN wp_deals_users_products up
ON u.ID = up.user_id
INNER JOIN $wpdb->usermeta um
ON u.ID = um.user_id            
WHERE up.pid = %d
AND  um.meta_key = '$capabilities' 
AND um.meta_value LIKE '%subscriber%'
";

echo $sql;

调试$ sql并尝试使用phpmyadmin获取的查询。

WPDB无法正常工作，SQL查询在phpmyadmin中工作

2 个答案: