Question

我试图使用函数与前方,但获取错误。这是我的代码:

function student_image_list(){
    global $host, $dbname, $user, $pass;
    $DBH = new PDO("mysql:host=$host;dbname=$dbname", $user, $pass);
    $STH = $DBH->query("SELECT DISTINCT folder_name from school_students_images");
    $STH->setFetchMode(PDO::FETCH_ASSOC);
    $row = $STH->fetch();
    return $row;
}

我要从表格 school_students_images 上打印所有唯一的文件夹名称, 我尝试这样做 :

$folder_name1 = student_image_list();
foreach($folder_name1[ folder_name ] as $folder_name)  
    { 
        echo   <div class="mws-panel grid_2">
               <div class="mws-panel-header">
               <span> .$folder_name. </span>
               </div>                            
               </div> ;
    }

Error Warning: Invalid argument supplied for foreach() in student_image_list_raw.php on line 41 error in this line

foreach($folder_name1[ folder_name ] as $folder_name)

请帮助。

Answer 1

您只抓取一行, 似乎没有文件夹名称密钥, 您必须

function student_image_list(){
    global $host, $dbname, $user, $pass;
    $DBH = new PDO("mysql:host=$host;dbname=$dbname", $user, $pass);
    $STH = $DBH->query("SELECT DISTINCT folder_name from school_students_images");
    $STH->setFetchMode(PDO::FETCH_ASSOC);
    return $STH;
}

$STH = student_image_list();
while (($row = $STH->fetch()) !== false) {
    echo   <div class="mws-panel grid_2">
           <div class="mws-panel-header">
           <span> .$row[ folder_name ]. </span>
           </div>                            
           </div> ;

}

友情链接