使用ajax将图像发送到PLAY框架

2 回答

• 0

  您可以像here一样使用Formdata

  var fd = new FormData();    
  fd.append( 'file', input.files[0] );
  
  $.ajax({
    url: 'http://example.com/script.php',
    data: fd,
    processData: false,
    contentType: false,
    type: 'POST',
    success: function(data){
      alert(data);
    }
  });
  

  回复于 2024-04-20T20:18:32+08:00
• 0

  也许这可以帮到你 . 这实际上是如何通过表单发送和存储mp3（base64）到PLAY Framework . 我希望这可以帮助你一点：

  main.scala.html

  <!-- AUDIO -->
      <section id="sound">
          <div class="container">
              <div class="row">
                  <div class="col-lg-12 text-center">
      <form action="**@routes.Application.upload()**"  name="addProductForm" id="addProductForm" enctype="multipart/form-data" method="post">
          <label>Select file.
              <input name="base64" type="text" accept="*/*" value="data:audio/mp3;base64,//sQxAAABKh...JZsdf2y/FOtQ==" >
              <!-- START This is actually not needed.Because you dont want use the uploader itself -->
              <input name="picture" type="file" accept="*/*" >
              <!-- ENDE This is actually not needed.Because you dont want use the uploader itself -->
            </label>
          <input name="submit" type="submit" size="50" value="senden">
  
      </form>
                  </div>
              </div>
          </div>
      </section>
      <!-- AUDIO -->
  

  非常重要，在Java中使用FileUtils只需安装org.apache.common.io.FileUtils（!!!）

  Application.java

  public static Result upload(){
  
      System.out.println("Entered the upload() method !!!");
  
      play.mvc.Http.MultipartFormData body = request().body().asMultipartFormData();
      DynamicForm dynamicForm = Form.form().bindFromRequest();
  
      play.mvc.Http.MultipartFormData.FilePart picture = body.getFile("picture");
      String str = dynamicForm.get("base64");
  
      // example: "data:audio/mp3;base64,//sQxAAABKhrG...f2y/FOtQ==";
  
      String[] strsplited = str.split(",");
      String b64 = strsplited[1];
      if (picture != null) {
  
          File file = picture.getFile();
          String fileName = file.getName();
  
          System.out.println("\n\n"+file.getPath()+"/"+fileName);
          try {
              /* START This is actually not needed.Because you dont want use the uploader itself */
              File destination = new File("/Applications/MAMP/htdocs/play_old/myapp/public/upload/", file.getName());
              FileUtils.moveFile(file,destination);
              /* ENDE This is actually not needed.Because you dont want use the uploader itself */
  
              java.security.SecureRandom random = new java.security.SecureRandom();
              String newFileName = new BigInteger(130, random).toString(32);
              byte[] data = Base64.getDecoder().decode(b64);
              File dest = new File("/Applications/MAMP/htdocs/play_old/myapp/public/upload/", newFileName+".mp3");
              FileUtils.writeByteArrayToFile(dest,data);
              System.out.println("Byte size: "+data.length);
              System.out.println("Create new mp3 file :"+bi+".mp3");
  
          } catch(IOException e){
              e.getMessage();
          }
  
      } else {
          flash("error", "Missing file");
          return badRequest();
      }
  
      return ok("File uploaded");
  
  }
  

  routes

  POST  /upload           controllers.Application.upload()
  

  回复于 2024-04-20T20:18:32+08:00