如何通过此代码传递文件/图像

Question

我想将图像和标题字段值传递给PHP，我通常使用$ _FILES数组直接处理文件上传，我不知道如何创建/使用ajax将此数组传递给PHP。我的形式：

<script> 
      $(document).ready(function() { 
       $('form').submit(function(event) { //Trigger on form submit 
        $('#name + .throw_error').empty(); //Clear the messages first 
        $('#success').empty(); 

        var guestbookSendMessage = { //Fetch form data 
         'name' : $('input[name=name]').val(), //Store name fields value 
         'msg': $('textarea[name=msg]').val() 

        }; 

        $.ajax({ //Process the form using $.ajax() 
         type  : 'POST', //Method type 
         url   : 'php/process.php', //Your form processing file url 
         data  : guestbookSendMessage, //Forms name 
         dataType : 'json', 
         success  : function(data) { 

         if (!data.success) { //If fails 
          if (data.errors.name) { //Returned if any error from process.php 
           $('.throw_error').fadeIn(1000).html(data.errors.name); //Throw relevant error 
          } 
         } else { 
           $('#success').fadeIn(1000).append('<p>' + data.posted + '</p>'); //If successful, than throw a success message 
          } 
         } 
        }); 
        event.preventDefault(); //Prevent the default submit 
       }); 
      }); 
     </script> 

Answer 1

试试这个：

的Jquery：

$(document).ready(function(){ 
    $('#upload').on('click', function() { 
     var file_data = $('#pic').prop('files')[0]; 
     var form_data = new FormData(); 
     form_data.append('file', file_data); 

     $.ajax({ 
       url   : 'upload.php',  // point to server-side PHP script 
       dataType : 'text',   // what to expect back from the PHP script, if anything 
       cache  : false, 
       contentType : false, 
       processData : false, 
       data  : form_data,       
       type  : 'post', 
       success  : function(output){ 
        alert(output);    // display response from the PHP script, if any 
       } 
     }); 
     $('#pic').val('');      /* Clear the file container */ 
    }); 
}); 

PHP的：

<?php 
    if ($_FILES['file']['error'] > 0){ 
     echo 'Error: ' . $_FILES['file']['error'] . '<br>'; 
    } 
    else { 
     if(move_uploaded_file($_FILES['file']['tmp_name'], 'uploads/' . $_FILES['file']['name'])) 
     { 
      echo "File Uploaded Successfully"; 
     } 
    } 

?> 

Answer 2

只是改变你的代码与这个jQuery代码，并尝试

var guestbooksendmessage = new FormData(); 
    guestbooksendmessage.append('file', input.files[0]); 
    guestbooksendmessage.append('name', $('input[name=name]').val()); 
    guestbooksendmessage.append('msg', $('textarea[name=msg]').val()); 
    $.ajax({ 

      type: "POST", 
        url: 'php/process.php', 
        dataType: "json", 
        data: guestbooksendmessage, 
        processData: false, 
        contentType: false, 
        cache: false, 
        success: function (data) { 
          alert(data) 
        } 
       }); 

试试这个代码