“Unggah file menggunakan AJAX” Kode Jawaban

Unggah file menggunakan AJAX

// add jquery
// <script src="https://code.jquery.com/jquery-3.6.0.min.js" integrity="sha256-/xUj+3OJU5yExlq6GSYGSHk7tPXikynS7ogEvDej/m4=" crossorigin="anonymous"></script>
$(document).ready(function() {
  $("#form").on('submit', (function(e) {
    e.preventDefault();
    $.ajax({
      url: $(this).attr('action'),
      type: "POST",
      data: new FormData(this),
      contentType: false,
      cache: false,
      processData: false,
      success: function(response) {
        $("#form").trigger("reset"); // to reset form input fields
      },
      error: function(e) {
        console.log(e);
      }
    });
  }));
});
Terrible Tern

unggahan file dalam php melalui jax

$('#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: '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(php_script_response){
            alert(php_script_response); // display response from the PHP script, if any
        }
     });
});
Dev

unggahan file dalam php melalui jax

**1. index.php**
<body>
    <span id="msg" style="color:red"></span><br/>
    <input type="file" id="photo"><br/>
  <script type="text/javascript" src="jquery-3.2.1.min.js"></script>
  <script type="text/javascript">
    $(document).ready(function(){
      $(document).on('change','#photo',function(){
        var property = document.getElementById('photo').files[0];
        var image_name = property.name;
        var image_extension = image_name.split('.').pop().toLowerCase();

        if(jQuery.inArray(image_extension,['gif','jpg','jpeg','']) == -1){
          alert("Invalid image file");
        }

        var form_data = new FormData();
        form_data.append("file",property);
        $.ajax({
          url:'upload.php',
          method:'POST',
          data:form_data,
          contentType:false,
          cache:false,
          processData:false,
          beforeSend:function(){
            $('#msg').html('Loading......');
          },
          success:function(data){
            console.log(data);
            $('#msg').html(data);
          }
        });
      });
    });
  </script>
</body>

**2.upload.php**
<?php
if($_FILES['file']['name'] != ''){
    $test = explode('.', $_FILES['file']['name']);
    $extension = end($test);    
    $name = rand(100,999).'.'.$extension;

    $location = 'uploads/'.$name;
    move_uploaded_file($_FILES['file']['tmp_name'], $location);

    echo '<img src="'.$location.'" height="100" width="100" />';
}
Dev

Unggah formulir dengan tipe DOC di AJAX

$(".submitbtn").on("click", function(e) {

    var form = $("#Form");

    // you can't pass Jquery form it has to be javascript form object
    var formData = new FormData(form[0]);

    //if you only need to upload files then 
    //Grab the File upload control and append each file manually to FormData
    //var files = form.find("#fileupload")[0].files;

    //$.each(files, function() {
    //  var file = $(this);
    //  formData.append(file[0].name, file[0]);
    //});

    if ($(form).valid()) {
        $.ajax({
            type: "POST",
            url: $(form).prop("action"),
            //dataType: 'json', //not sure but works for me without this
            data: formData,
            contentType: false, //this is requireded please see answers above
            processData: false, //this is requireded please see answers above
            //cache: false, //not sure but works for me without this
            error   : ErrorHandler,
            success : successHandler
        });
    }
});
Thoughtful Tapir

unggahan file dalam php melalui jax

var formData = new FormData($("#YOUR_FORM_ID")[0]);
$.ajax({
    url: "upload.php",
    type: "POST",
    data : formData,
    processData: false,
    contentType: false,
    beforeSend: function() {

    },
    success: function(data){




    },
    error: function(xhr, ajaxOptions, thrownError) {
       console.log(thrownError + "\r\n" + xhr.statusText + "\r\n" + xhr.responseText);
    }
});
Dev

unggahan file dalam php melalui jax

<?
$data = array();
    //check with your logic
    if (isset($_FILES)) {
        $error = false;
        $files = array();

        $uploaddir = $target_dir;
        foreach ($_FILES as $file) {
            if (move_uploaded_file($file['tmp_name'], $uploaddir . basename( $file['name']))) {
                $files[] = $uploaddir . $file['name'];
            } else {
                $error = true;
            }
        }
        $data = ($error) ? array('error' => 'There was an error uploading your files') : array('files' => $files);
    } else {
        $data = array('success' => 'NO FILES ARE SENT','formData' => $_REQUEST);
    }

    echo json_encode($data);
?>
Dev

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