Question

我有 wp_places 自定义表格，当我打印数组时，我得到了这个：

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

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

现在我想以这种方式破坏价值（24,29）但是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);

有任何想法或建议吗？感谢。

Answer 1

$wpdb->get_results()已为您抓取，您无需拨打mysql_fetch_array

鉴于您想要做什么，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

这是一个包含结果的PHP对象，它不是MySQL结果。

查看docs，应该像

一样使用

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

Wordpress MySQL结果资源无效

2 个答案: