Question

大家好我已经制作了一个图库删除功能，当我开始工作时，我将适应图片报告功能（之后）但是当传递ajax时，功能dosnt在这里做任何事情都是一些代码片段......

用于触发功能的链接

$gallery_list .= '

<div class="pic_holder" onclick="DeletePhoto(\''.$filename.'\')">
<img class="pics" src="user/'.$u.'/'.$filename.'"></a></div>

';

功能

function DeletePhoto(filename) {
    var conf = confirm("Press OK to delete this photo.");
    if (conf != true) {
        return false;
    }
    _("deletelink").style.visibility = "hidden";
    var ajax = ajaxObj("POST", "php_parsers/photo_system.php");
    ajax.onreadystatechange = function() {
        if (ajaxReturn(ajax) == true) {
            if (ajax.responseText == "deleted_ok") {
                alert("This picture has been deleted successfully. We will now refresh the page for you.");
                window.location = "photos.php?u=<?php echo $u; ?>";
            }
        }
    }
    ajax.send("delete=photo&filename=" + filename);
}

和ajax解析器代码

   if (isset($_POST["delete"]) && $_POST["filename"] != "") {
       $filename1 = preg_replace('#[^a-z0-9]#i', '', $_POST["filename"]);
       $query = mysqli_query($db_conx, "SELECT user, filename FROM photos WHERE filename='$filename1' LIMIT 1");
       $row = mysqli_fetch_row($query);
       $user = $row[0];
       $filename = $row[1];
       if ($user == $log_username) {
           $picurl = "../user/$log_username/$filename";
           if (file_exists($picurl)) {
               if ($filename != "avatardefault.jpg") {
                   unlink($picurl);
               }
               $sql = "DELETE FROM photos WHERE filename='$filename1' LIMIT 1";
               $query = mysqli_query($db_conx, $sql);
           }
       }
       mysqli_close($db_conx);
       echo "deleted_ok";
       exit();
   }

任何想法我都做错了什么？因为当我点击按钮时，我得到了确认框，但是当我点击OK时，没有任何反应，ps我真的不希望编辑触发该功能的链接，因为该部分有效，我觉得错误是在其中一个最后两个代码片段。

为什么这个ajax php解析函数失败了

0 个答案: