mysql - Wordpress MySQL 结果资源无效

Question

我有wp_places自定义表，并且在打印数组时得到了这个：

[0] => stdClass Object
        (
            [home_location] => 24
        )

[1] => stdClass Object
        (
            [home_location] => 29
        )

现在我想以这种方式（24,29）内爆值，但在我的代码中我收到了这个错误：

<b>Warning</b>:  mysql_fetch_array(): supplied argument is not a valid MySQL result resource

我的代码

$getGroupType = $_POST['parent_category'];
    $result = $wpdb->get_results( "SELECT home_location FROM wp_places WHERE blood_group LIKE '".$getGroupType."%'" );


    $bgroup = Array();
    while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
        $bgroup[] =  implode(',',$row);
    }
    echo implode(',',$bgroup);

有什么想法或建议吗？谢谢。

Answer 1

$wpdb->get_results()已经为您完成了抓取，您无需致电mysql_fetch_array

鉴于您想做的事情，您的代码应如下所示：

$getGroupType = $_POST['parent_category'];
$result = $wpdb->get_results( "SELECT home_location FROM wp_places WHERE blood_group LIKE '".$getGroupType."%'" );


$bgroup = Array();
foreach ($result as $location) {
    $bgroup[] =  $location->home_location;
}
echo '('.implode(',',$bgroup).')';

Answer 2

它是一个包含结果的 PHP 对象，而不是 MySQL 结果。

查看文档，它应该像这样使用

foreach ($result as $row) {
    $bgroup[] = $row->home_location;
}
echo implode(',',$bgroup)