Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
575 views
in Technique[技术] by (71.8m points)

javascript - How to send data to server while upload file?

When a user uploads a picture to a server, the picture must to be stored like pkUsre.extention, exemple 1.jpg, where the number 1 is the users primary key. I have the follow code that works fine to upload the file, but I dont know how to send the pkUser to server with the picture. Any idea how to do that?

//Html code
   <form id="form1" enctype="multipart/form-data" method="post" >
    <div class="row">
      <input type="file" name="fileToUpload" id="fileToUpload" onchange="fileSelected();"/>
    </div>
    <div id="fileName"></div>
    <div id="fileSize"></div>
    <div id="fileType"></div>
    <div class="row">
      <input type="button" id="btnSd"  value="Upload" />
    </div>
    <div id="progressNumber"></div>
  </form>

//PHP code
    if ( move_uploaded_file( $_FILES['fileToUpload']['tmp_name'], "pic/".      basename($_FILES['fileToUpload']['name']) ) ) {
        echo basename($_FILES['fileToUpload']['name']);
    }
    else {
        echo "There was a problem uploading your file - please try again.";
    }   

//Javascript code
    <script type="text/javascript">
      function fileSelected() {
        var file = document.getElementById('fileToUpload').files[0];
        if (file) {
          var fileSize = 0;
          if (file.size > 1024 * 1024)
            fileSize = (Math.round(file.size * 100 / (1024 * 1024)) / 100).toString() + 'MB';
          else
            fileSize = (Math.round(file.size * 100 / 1024) / 100).toString() + 'KB';

          document.getElementById('fileName').innerHTML = 'Name: ' + file.name;
          document.getElementById('fileSize').innerHTML = 'Size: ' + fileSize;
          document.getElementById('fileType').innerHTML = 'Type: ' + file.type;
        }
      }

      function uploadFile() {
        var fd = new FormData();
        fd.append("fileToUpload", document.getElementById('fileToUpload').files[0]);
        var xhr = new XMLHttpRequest();
        xhr.upload.addEventListener("progress", uploadProgress, false);
        xhr.addEventListener("load", uploadComplete, false);
        xhr.addEventListener("error", uploadFailed, false);
        xhr.addEventListener("abort", uploadCanceled, false);
        xhr.open("POST", "uploadFoto.php");
        xhr.send(fd);
      }

      function uploadProgress(evt) {
        if (evt.lengthComputable) {
          var percentComplete = Math.round(evt.loaded * 100 / evt.total);
          document.getElementById('progressNumber').innerHTML = percentComplete.toString() + '%';
        }
        else {
          document.getElementById('progressNumber').innerHTML = 'unable to compute';
        }
      }

      function uploadComplete(evt) {
        /* This event is raised when the server send back a response */
        alert(evt.target.responseText);
      }

      function uploadFailed(evt) {
        alert("There was an error attempting to upload the file.");
      }

      function uploadCanceled(evt) {
        alert("The upload has been canceled by the user or the browser dropped the connection.");
      }
    </script> 
   <script type="text/javascript">
       $(document).on("click", "#btnSd", function(){
          uploadFile();      
       });     
   </script>
See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)
    only file uplod name in server database and move to file in folder user_post.
    file copy in filder and only name is database save.


    function getuniqkey($special='')
        {
            return md5(date("Y-m-d H:i:s").uniqid(rand(), true).time().$special);
        }
    if(isset($_FILES["fileToUpload"]))
        {
            $tmp_name = $_FILES["fileToUpload"]["tmp_name"];
            $name1 = $_FILES["fileToUpload"]["name"];       
            $datasheet_new_name=getuniqkey($tmp_name).substr($name1,strrpos($name1,"."));
            if(copy($_FILES["fileToUpload"]["tmp_name"], "user_post/".$datasheet_new_name))
            {
                $file_name = "http://domain name.com/user_post/{$datasheet_new_name}";
            }
            else
            {   
                echo'user file not upload.';


            }

    }        
    echo $file_name;

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

1.4m articles

1.4m replys

5 comments

57.0k users

...