将输入文件放入FormData jquery以提交给PHP

Question

我知道这个问题已经有问题,但我还是不能正确地做到这一点.需要一些帮助.

我需要上传一个带有附加数据的文件.

我的input.php:

<input type="file"  id="foto_path" name="foto_path"  />
<input type="button" value="Add" onclick="javascript:sendForm()" />

我用javascript发送:

function sendForm() {
    var fileInput = document.querySelector('#foto_path');
    var oMyForm = new FormData();
    var nip=123223374;//it will be generated by php, for temporary i just hardcode it
    oMyForm.append("foto_path", fileInput);
    oMyForm.append("nip",nip );
    var oReq = new XMLHttpRequest();
    oReq.open("POST", "upload-file.php", true);
    oReq.onload = function(oEvent) {
        if (oReq.status == 200) {
          //oOutput.innerHTML = "Uploaded!";
          alert('success');
        } else {
          //oOutput.innerHTML = "Error " + oReq.status + " occurred uploading your file.<br \/>";
          alert('failed');
        }
      };
    oReq.send(oMyForm);
}

当我发送到upload-file.php时:

logapp("post -> ".print_r($_POST,true));//logapp is just function to log to file
logapp("files -> ".print_r($_FILES,true));

我从日志中得到了这个:

09/07/2013 02:47:06 pm :: post -> Array
(
    [foto_path] => [object HTMLImageElement]
    [nip] => 123223374
)

09/07/2013 02:47:06 pm :: files -> Array
(

)

我得到了成功的警报,但我需要得到文件foto_path的$_FILES不$_POST.我的问题是为什么检测到的输入文件$_POST["foto_path"]不是$_FILES["foto_path"].如何更改,$_FILES["foto_path"]以便我可以开始使用上传的文件？

Answer 1

更新:我终于解决了这个问题

var ft=$('#foto_path_upload')[0].files[0];
oMyForm.append("foto_path_upload", ft);

这是我从日志中得到的:

09/07/2013 04:26:53 pm :: files -> Array
(
    [foto_path_upload] => Array
        (
           [name] => noimages.jpg
           [type] => image/jpeg
           [tmp_name] => D:\xampp\tmp\php4E90.tmp
           [error] => 0
           [size] => 3642
         )

 )    

问题解决了.大家好