46

我设计了一个简单的表单，允许用户将文件上传到服务器。最初，表单包含一个“浏览”按钮。如果用户想要上传多个文件，他需要点击“添加更多文件”按钮，该按钮会在表单中添加另一个“浏览”按钮。提交表单时，文件上传过程在“upload.php”文件中处理。它非常适合上传多个文件。现在我需要使用 jQuery 的 '.submit()' 提交表单并向 'upload.php' 文件发送 ajax ['.ajax()'] 请求以处理文件上传。

这是我的 HTML 表单：

<form enctype="multipart/form-data" action="upload.php" method="post">
    <input name="file[]" type="file" />
    <button class="add_more">Add More Files</button>
    <input type="button" id="upload" value="Upload File" />
</form>

这是JavaScript：

$(document).ready(function(){
    $('.add_more').click(function(e){
        e.preventDefault();
        $(this).before("<input name='file[]' type='file' />");
    });
});

这是处理文件上传的代码：

for($i=0; $i<count($_FILES['file']['name']); $i++){
$target_path = "uploads/";
$ext = explode('.', basename( $_FILES['file']['name'][$i]));
$target_path = $target_path . md5(uniqid()) . "." . $ext[count($ext)-1]; 

if(move_uploaded_file($_FILES['file']['tmp_name'][$i], $target_path)) {
    echo "The file has been uploaded successfully <br />";
} else{
    echo "There was an error uploading the file, please try again! <br />";
}

}

任何关于我应该如何编写我的 '.submit()' 函数的建议都会非常有帮助。

4

4 回答 4

60

最后，我通过使用以下代码找到了解决方案：

$('body').on('click', '#upload', function(e){
        e.preventDefault();
        var formData = new FormData($(this).parents('form')[0]);

        $.ajax({
            url: 'upload.php',
            type: 'POST',
            xhr: function() {
                var myXhr = $.ajaxSettings.xhr();
                return myXhr;
            },
            success: function (data) {
                alert("Data Uploaded: "+data);
            },
            data: formData,
            cache: false,
            contentType: false,
            processData: false
        });
        return false;
});
于 2013-10-28T11:35:12.843 回答
51

HTML

<form enctype="multipart/form-data" action="upload.php" method="post">
    <input name="file[]" type="file" />
    <button class="add_more">Add More Files</button>
    <input type="button" value="Upload File" id="upload"/>
</form>

Javascript

 $(document).ready(function(){
    $('.add_more').click(function(e){
        e.preventDefault();
        $(this).before("<input name='file[]' type='file'/>");
    });
});

用于ajax上传

$('#upload').click(function() {
    var filedata = document.getElementsByName("file"),
            formdata = false;
    if (window.FormData) {
        formdata = new FormData();
    }
    var i = 0, len = filedata.files.length, img, reader, file;

    for (; i < len; i++) {
        file = filedata.files[i];

        if (window.FileReader) {
            reader = new FileReader();
            reader.onloadend = function(e) {
                showUploadedItem(e.target.result, file.fileName);
            };
            reader.readAsDataURL(file);
        }
        if (formdata) {
            formdata.append("file", file);
        }
    }
    if (formdata) {
        $.ajax({
            url: "/path to upload/",
            type: "POST",
            data: formdata,
            processData: false,
            contentType: false,
            success: function(res) {

            },       
            error: function(res) {

             }       
             });
            }
        });

PHP

for($i=0; $i<count($_FILES['file']['name']); $i++){
    $target_path = "uploads/";
    $ext = explode('.', basename( $_FILES['file']['name'][$i]));
    $target_path = $target_path . md5(uniqid()) . "." . $ext[count($ext)-1]; 

    if(move_uploaded_file($_FILES['file']['tmp_name'][$i], $target_path)) {
        echo "The file has been uploaded successfully <br />";
    } else{
        echo "There was an error uploading the file, please try again! <br />";
    }
}

/** 
    Edit: $target_path variable need to be reinitialized and should 
    be inside for loop to avoid appending previous file name to new one. 
*/

请使用上面的脚本进行 ajax 上传。它会工作

于 2013-10-10T12:39:59.490 回答
4

使用此源代码，您可以通过 ajax 一个一个地上传多个文件，如 google。您还可以看到上传进度

HTML

 <input type="file" id="multiupload" name="uploadFiledd[]" multiple >
 <button type="button" id="upcvr" class="btn btn-primary">Start Upload</button>
 <div id="uploadsts"></div>

Javascript

    <script>

    function uploadajax(ttl,cl){

    var fileList = $('#multiupload').prop("files");
    $('#prog'+cl).removeClass('loading-prep').addClass('upload-image');

    var form_data =  "";

    form_data = new FormData();
    form_data.append("upload_image", fileList[cl]);


    var request = $.ajax({
              url: "upload.php",
              cache: false,
              contentType: false,
              processData: false,
              async: true,
              data: form_data,
              type: 'POST', 
              xhr: function() {  
                  var xhr = $.ajaxSettings.xhr();
                  if(xhr.upload){ 
                  xhr.upload.addEventListener('progress', function(event){
                      var percent = 0;
                      if (event.lengthComputable) {
                          percent = Math.ceil(event.loaded / event.total * 100);
                      }
                      $('#prog'+cl).text(percent+'%') 
                   }, false);
                 }
                 return xhr;
              },
              success: function (res, status) {
                  if (status == 'success') {
                      percent = 0;
                      $('#prog' + cl).text('');
                      $('#prog' + cl).text('--Success: ');
                      if (cl < ttl) {
                          uploadajax(ttl, cl + 1);
                      } else {
                          alert('Done');
                      }
                  }
              },
              fail: function (res) {
                  alert('Failed');
              }    
          })
    }

    $('#upcvr').click(function(){
        var fileList = $('#multiupload').prop("files");
        $('#uploadsts').html('');
        var i;
        for ( i = 0; i < fileList.length; i++) {
            $('#uploadsts').append('<p class="upload-page">'+fileList[i].name+'<span class="loading-prep" id="prog'+i+'"></span></p>');
            if(i == fileList.length-1){
                uploadajax(fileList.length-1,0);
            }
         }
    });
    </script>

PHP

upload.php
    move_uploaded_file($_FILES["upload_image"]["tmp_name"],$_FILES["upload_image"]["name"]);
于 2018-08-04T10:50:51.267 回答
4

我的解决方案

  • 假设表单 id = "my_form_id"
  • 它从 HTML中检测表单方法和表单操作

jQuery代码

$('#my_form_id').on('submit', function(e) {
    e.preventDefault();
    var formData = new FormData($(this)[0]);
    var msg_error = 'An error has occured. Please try again later.';
    var msg_timeout = 'The server is not responding';
    var message = '';
    var form = $('#my_form_id');
    $.ajax({
        data: formData,
        async: false,
        cache: false,
        processData: false,
        contentType: false,
        url: form.attr('action'),
        type: form.attr('method'),
        error: function(xhr, status, error) {
            if (status==="timeout") {
                alert(msg_timeout);
            } else {
                alert(msg_error);
            }
        },
        success: function(response) {
            alert(response);
        },
        timeout: 7000
    });
});
于 2016-05-05T11:49:47.780 回答