/PHP/Determine the percentage of the file uploaded to the server using php

Determine the percentage of the file uploaded to the server using php

$(function() {

    var bar = $('.bar');
    var percent = $('.percent');
    var status = $('#status');

    $('form').ajaxForm({
        beforeSend: function() {
            status.empty();
            var percentVal = '0%';
            bar.width(percentVal);
            percent.html(percentVal);
        },
        uploadProgress: function(event, position, total, percentComplete) {
            var percentVal = percentComplete + '%';
            bar.width(percentVal);
            percent.html(percentVal);
        },
        complete: function(xhr) {
            status.html(xhr.responseText);
        }
    });
}); 





<form action="file-echo2.php" method="post" enctype="multipart/form-data">
    <input type="file" name="myfile"><br>
    <input type="submit" value="Upload File to Server">
</form>

<div class="progress">
    <div class="bar"></div >
    <div class="percent">0%</div >
</div>

<div id="status"></div>

Source: stackoverflow.com

Determine the percentage of the file uploaded to the server using php

$.ajax({
  xhr: function() {
    var xhr = new window.XMLHttpRequest();

    xhr.upload.addEventListener("progress", function(evt) {
      if (evt.lengthComputable) {
        var percentComplete = evt.loaded / evt.total;
        percentComplete = parseInt(percentComplete * 100);
        console.log(percentComplete);

        if (percentComplete === 100) {

        }

      }
    }, false);

    return xhr;
  },
  url: posturlfile,
  type: "POST",
  data: JSON.stringify(fileuploaddata),
  contentType: "application/json",
  dataType: "json",
  success: function(result) {
    console.log(result);
  }
});

Source: stackoverflow.com

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