2013-07-01 42 views
1

wp_places定製表,我得到這個當我打印陣列:WordPress的MySQL的結果資源無效

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

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

現在我想爆這樣的方式(24,29)的價值,但在我的代碼我收到此錯誤:

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

我的代碼

$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); 

任何意見或建議?謝謝。

回答

5

$wpdb->get_results()已經做的獲取你,你不需要調用mysql_fetch_array

給你想要做什麼,你的代碼應該是這樣的:

$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).')'; 
0

它是包含一個結果PHP對象,它不是一個MySQL的結果。

望着docs,應該使用像

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