AJAX POST 파일 업로드 제어 PHP 파일

Question

AJAX 사용 PHP 업로드 컨트롤 값을 PHP 파일에 게시하고 있습니다. 아래

파일 업로드 컨트롤 :

<input class="form-control" type="file" id="policy_image" name="policy_image" accept="image/*"> 

AJAX 전화는 아래 :

$name = $_POST['policy']; 

$imgFile = $_FILES['file']['name']; 
$tmp_dir = $_FILES['file']['tmp_name']; 
$imgSize = $_FILES['file']['size']; 


    $upload_dir = 'assets/img/policy/'; // upload directory 

    $imgExt = strtolower(pathinfo($imgFile,PATHINFO_EXTENSION)); // get image extension 

    // valid image extensions 
    $valid_extensions = array('jpeg', 'jpg', 'png', 'gif'); // valid extensions 

    // rename uploading image 
    $userpic = rand(1000,1000000).".".$imgExt; 

    // allow valid image file formats 
    if(in_array($imgExt, $valid_extensions)){  
    // Check file size '5MB' 
    if($imgSize < 5000000)  { 
     move_uploaded_file($tmp_dir,$upload_dir.$userpic); 
    } 
    else{ 
     $errMSG = "Sorry, your file is too large."; 
    } 
    } 
    else{ 
    $errMSG = "Sorry, only JPG, JPEG, PNG & GIF files are allowed.";  
    } 


// if no error occured, continue .... 
if(!isset($errMSG)) 
{ 
    $stmt = $DB_con->prepare('INSERT INTO policy(name,image) VALUES(:name, :image)'); 
    $stmt->bindParam(':name',$name); 
    $stmt->bindParam(':image',$userpic); 

    if($stmt->execute()) 
    { 
    $successMSG = "new record succesfully inserted ..."; 
    } 
    else 
    { 
    $errMSG = "error while inserting...."; 
    } 
} 

: 내 PHP 파일에서

var file_data = $('#policy_image').prop('files')[0]; 
var policy_image = new FormData();     
policy_image.append('file', file_data); 

$.ajax({ 
    type: 'post', 
    url: 'addpolicy.php', 
    data: { 
     'policy': $('#policy').val(), 
     'policy_image': policy_image 
    },     
    enctype: 'multipart/form-data', 
    processData: false, 
    contentType: false,    
    success: function(data) { 
     alert('Success'); 
    } 
}); 

나는

아래 PHP 파일처럼 게시 된 값을 받고

이제 알겠습니다. Undefined index: policy 및 Undefined index: policy_image

Answer 1

같은 팅 오류이 시도 :

var file_data = $('#policy_image').prop('files')[0]; 
var form_data = new FormData(); 
form_data.append('policy_image', file_data); 

$.ajax({ 
    url   : 'addpolicy.php', 
    cache  : false, 
    contentType : false, 
    processData : false, 
    data  : form_data,       
    type  : 'post', 
    success  : function(output){ 
     alert(output); 
    } 
}); 

PHP :

당신은`file` 파일을 명명 된 FormData에서

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

Answer 2

upload.php 
     $('#upload').on('click', function() { 
      var file_data = $('#sortpicture').prop('files')[0]; 
      var form_data = new FormData();     
      form_data.append('file', file_data); 
      alert(form_data);        
      $.ajax({ 
         url: 'addpolicy.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(response){ 
          alert(response); 
         } 
      }); 
     }); 
    addpolicy.php 
    <?php 

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

    ?>