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从chrome扩展程序上传到php $ _FILE

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我正试图通过POST上的图像上传到javascript到我无法修改源代码的网站 .

该页面有一个允许您上传图像的表单:
<form enctype="multipart/form-data" action="/u.php" method="post"> <input name="file" type="file"> <input type="submit" value="Upload File"> </form>

我希望能够使用javascript上传图像,但我无法得到任何工作,我不确定这是否可能...

我的JS到目前为止:

file = document.getElementById('fileinput').files[0];
r = new FileReader();
r.onloadend = doUpload;
r.readAsBinaryString(file)

function doUpload(el){
    file = el.target.result;
XMLHttpRequest.prototype.sendAsBinary = function(string) {
    var bytes = Array.prototype.map.call(string, function(c) {
      return c.charCodeAt(0) & 0xff;
    });
    this.send(new Uint8Array(bytes).buffer);
  };
        var xhr = new XMLHttpRequest();
  xhr.open('POST', 'http://upload.domain.com/u.php', true);
  var boundary = 'ohaiimaboundary';
  xhr.setRequestHeader(
    'Content-Type', 'multipart/form-data; boundary=' + boundary);
  xhr.sendAsBinary([
    '--' + boundary,
    'Content-Disposition: form-data; name="file"; filename="test.jpg"',
    'Content-Type: multipart/form-data',
    '',
    file,
    '--' + boundary + '--'
  ].join('\r\n'));

}

谢谢

编辑:想出这个,有点,这应该工作一点点修改(png硬编码)

function doUpload(fl){
    var file = fl.target.result;
    XMLHttpRequest.prototype.sendAsBinary = function(datastr) {
            var bb = new BlobBuilder();
            var data = new ArrayBuffer(1);
            var ui8a = new Uint8Array(data, 0);
            for (var i in datastr) {
                    if (datastr.hasOwnProperty(i)) {
                            var chr = datastr[i];
                            var charcode = chr.charCodeAt(0)
                            var lowbyte = (charcode & 0xff)
                            ui8a[0] = lowbyte;
                            bb.append(data);
                    }
            }
            var blob = bb.getBlob();
            this.send(blob);
    }
    var xh = new XMLHttpRequest();
    xh.open('post', 'http://upload.domain.com/u.php', true);
    xh.onreadystatechange = function(){
        if(this.readyState != 4){
            return;
        }
        else{
            console.log(this.responseText);
        }
    };
    var boundary = '--fgsfds--';
    xh.setRequestHeader('Content-Type', 'multipart/form-data; boundary=' + boundary);
    xh.sendAsBinary([
        '--' + boundary,
        'Content-Type: image/png',
        'Content-Disposition: form-data; name="file"; filename="testz.png"',
        '',
        file,
        '--' + boundary + '--',
        ''].join('\n'));
}   
function mkUpload(){
    var r = new FileReader();
    r.onloadend = doUpload;
    r.readAsBinaryString(document.upload.file.files[0]);
}

测试PHP:

<?
echo sprintf('<pre>%s</pre>', print_r($_FILES, true));
?>

1 回答

  • 1

    其实,你试过 xhr.send(FormData) 吗? FormData允许您附加File对象,这些对象将被视为二进制内容,并与XHR一起使用它发送multipart / form-data . 没有必要自己构建它 .

    var formData = new FormData();
    
    var file = document.querySelector('input[type="file"]').files[0];
    if (file) {
      formData.append("file", file);
    }
    
    var xhr = new XMLHttpRequest();
    xhr.open('POST', '/u.php', true);
    xhr.onload = function(e) { ... };
    xhr.send(formData);
    

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