Wordpress MySQL结果资源无效
我有wp\u位置自定义表,当我打印数组时,我会得到这个:Wordpress MySQL结果资源无效,mysql,wordpress,while-loop,Mysql,Wordpress,While Loop,我有wp\u位置自定义表,当我打印数组时,我会得到这个: [0] => stdClass Object ( [home_location] => 24 ) [1] => stdClass Object ( [home_location] => 29 ) 现在我想这样内爆值(24,29),但在我的代码中,我得到了以下错误: <b>Warning&l
[0] => stdClass Object
(
[home_location] => 24
)
[1] => stdClass Object
(
[home_location] => 29
)
现在我想这样内爆值(24,29),但在我的代码中,我得到了以下错误:
<b>Warning</b>: mysql_fetch_array(): supplied argument is not a valid MySQL result resource
有什么想法或建议吗?谢谢。这是一个包含结果的PHP对象,不是MySQL结果 看看这个,应该这样使用
foreach ($result as $row) {
$bgroup[] = $row->home_location;
}
echo implode(',',$bgroup)
它是一个包含结果的PHP对象,而不是MySQL结果 看看这个,应该这样使用
foreach ($result as $row) {
$bgroup[] = $row->home_location;
}
echo implode(',',$bgroup)
$wpdb->get_results()
已经为您进行了抓取,您不需要调用mysql_fetch_array
考虑到您要执行的操作,您的代码应该如下所示:
$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).')';
$wpdb->get_results()
已经为您进行了抓取,您不需要调用mysql_fetch_array
考虑到您要执行的操作,您的代码应该如下所示:
$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).')';