通过Ajax传递Blob以生成文件

Question

我正在尝试捕获audiorecorder（https://github.com/cwilso/AudioRecorder）并通过Ajax向Blob发送一个php文件，该文件将接收Blob内容并创建文件（在这种情况下为wave文件）。

Ajax呼叫：

audioRecorder.exportWAV(function(blob) {
      var url = (window.URL || window.webkitURL).createObjectURL(blob);
      console.log(url);
      var filename = <?php echo $filename;?>;
      $.ajaxFileUpload({
        url :  "lib/vocal_render.php",
        secureuri      :false,
        dataType : blob.type,
        data: blob,
        success: function(data, status) {
          if(data.status != 'error')
            alert("boa!");
        }
      });
    });

和我的php文件（vocal_render.php）：

<?php

if(!empty($_POST)){
    $data = implode($_POST); //transforms the char array with the blob url to a string
    $fname = "11" . ".wav";

    $file = fopen("../ext/wav/testes/" .$fname, 'w');
    fwrite($file, $data);
    fclose($file);
}?>

PS：我是blob和ajax的新手。提前致谢。

Answer 1

尝试将文件上传为表单数据

audioRecorder.exportWAV(function(blob) {

      var url = (window.URL || window.webkitURL).createObjectURL(blob);
      console.log(url);

      var filename = <?php echo $filename;?>;
      var data = new FormData();
      data.append('file', blob);

      $.ajax({
        url :  "lib/vocal_render.php",
        type: 'POST',
        data: data,
        contentType: false,
        processData: false,
        success: function(data) {
          alert("boa!");
        },    
        error: function() {
          alert("not so boa!");
        }
      });
});

。

<?php

if(isset($_FILES['file']) and !$_FILES['file']['error']){
    $fname = "11" . ".wav";

    move_uploaded_file($_FILES['file']['tmp_name'], "../ext/wav/testes/" . $fname);
}
?>