javascript - 使用ajax形式发布数据id

Question

我有一个关于 ajax 表单的问题。所以，你可以看到我下面的ajax代码有 data-id 。我在 .pmsc div 中获取此 id。但是当我按 #upmsc 上传图像时，我没有收到任何错误，只是从 chrome 开发人员控制台获取此结果，没有上传，也没有在数据库中插入图像名称。我在这里做错了什么。

然后ajax提交表单是这样的

$('body').on('change', '#upmsc', function(e) {
   e.preventDefault();
   var id = $(".pmsc").attr("data-id");
   var data = 'id=' + id;
   $("#musicform").ajaxForm({
      type: "POST",
      data: data,
      cache: false,
      target: '.appended',
      beforeSubmit: function() {
         // Do Something
      },
      success: function(response) {
         // Do Something
         $(".showecho").html(response);
      },
      error: function() {}
   }).submit();
});

我的 html 表单是这样的:

    <form id="cfrm" class="options-form" method="post" enctype="multipart/form-data" action="'.$base_url.'upload.php">
       <label class="mcup" for="upcmsc"></label>
<input type="file" name="musicCover" id="upcmsc" />
    </form>
    <div class="pmsc" id="56">
    </div>

PHP 代码是这样的

<?php
include_once 'inc/inc.php';
$valid_formats = array("jpg", "png", "gif", "bmp","jpeg","PNG","JPG","JPEG","GIF","BMP");
if(isset($_POST) and $_SERVER['REQUEST_METHOD'] == "POST" && isset($_POST['id'])) {
    $id = mysqli_real_escape_string($db, $_POST['id']);
    $name = $_FILES['musicCover']['name'];
    $size = $_FILES['musicCover']['size'];
    if(strlen($name)) {
        $ext = getExtension($name);
        if(in_array($ext,$valid_formats)) {
            if($size<(50024*50024)) { 
                $actual_image_name = time().$uid.".".$ext;
                $tmp = $_FILES['musicCover']['tmp_name'];
                if(move_uploaded_file($tmp, $mcoverpath.$actual_image_name)) {
                    $data=$HuHu->Music_Cover_Image_Upload($id,$actual_image_name);
                    if($data){
                        echo '<img class="cover covermsc" src="'.$base_url.$mcoverpath.$actual_image_name.'">';
                        }
                  } else {
                      echo "Fail upload folder with read access.";
                }
                } else
                  echo "Image file size max 1 MB";                  
       } else
          echo "Invalid file format.";
     } else
         echo "Please select image..!";
        exit;
 }


?>

Answer 1

var data = 'id=' + id;
$("#musicform").ajaxForm({
  type: "POST",
  data: data,

您将附加数据设置为字符串，而它应该是对象。那么 PHP 代码中的检查就会失败。所以试试这个:

var data = {id: id};

发现 PHP 代码大括号还有一些问题，试试这个:

<?php
include_once 'inc/inc.php';
$valid_formats = array("jpg", "png", "gif", "bmp","jpeg","PNG","JPG","JPEG","GIF","BMP");
if(isset($_POST) and $_SERVER['REQUEST_METHOD'] == "POST" && isset($_POST['id'])) {
    $id = mysqli_real_escape_string($db, $_POST['id']);
    $name = $_FILES['musicCover']['name'];
    $size = $_FILES['musicCover']['size'];

    if(strlen($name)) {
        $ext = getExtension($name);
        if(in_array($ext,$valid_formats)) {
            if($size<(50024*50024)) { 
                $actual_image_name = time().$uid.".".$ext;
                $tmp = $_FILES['musicCover']['tmp_name'];

                if(move_uploaded_file($tmp, $mcoverpath.$actual_image_name)) {
                    $data=$HuHu->Music_Cover_Image_Upload($id,$actual_image_name);
                    if($newdata){
                        echo '<img class="cover covermsc" src="'.$base_url.$mcoverpath.$actual_image_name.'">';
                    }
                } else {
                    echo "Fail upload folder with read access.";
                }
            } else {
                echo "Image file size max 1 MB";        
            }
        } else {
            echo "Invalid file format.";
        }
    } else {
        echo "Please select image..!";
        exit;
    }
} else {
    echo '1';   
}
?>