2013-05-03 104 views
7

我有以下代碼:上傳中遇到問題的blob直接到S3

var fd = new FormData(); 

     var key = "events/" + (new Date).getTime() + '-'; 

     fd.append('key', key); 
     fd.append('acl', Acl); 
     fd.append('Content-Type', "image/jpeg"); 
     fd.append('AWSAccessKeyId', AWSAccessKeyId); 
     fd.append('policy', Policy); 
     fd.append('name', "Policy13492345"); 
     fd.append('success_action_status', "201"); 
     fd.append('signature', Signature);   
     fd.append("file", blob); 
     fd.append("filename", fileName + ".jpg"); 
     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', 'https://s3.amazonaws.com/' + Bucket + '/', true); 

     xhr.send(fd); 

當這個請求經過我收到以下錯誤:

<Code>AccessDenied</Code><Message>Invalid according to Policy: Policy Condition failed: ["starts-with", "$Filename", ""]</Message>

我不知道我是什麼做錯了,我生成我的blob像這樣:

function dataURItoBlob(dataURI) { 
       var binary = atob(dataURI.split(',')[1]); 
       var array = []; 
       for (var i = 0; i < binary.length; i++) { 
        array.push(binary.charCodeAt(i)); 
       } 
       var mimeString = dataURI.split(',')[0].split(':')[1].split(';')[0] 
       return new Blob([new Uint8Array(array)], { type: mimeString }); 
      } 

這是米y要求:

------WebKitFormBoundaryxh8thnHAmDhZQuXE 
Content-Disposition: form-data; name="key" 

events/1367541109750- 
------WebKitFormBoundaryxh8thnHAmDhZQuXE 
Content-Disposition: form-data; name="acl" 

private 
------WebKitFormBoundaryxh8thnHAmDhZQuXE 
Content-Disposition: form-data; name="Content-Type" 

image/jpeg 
------WebKitFormBoundaryxh8thnHAmDhZQuXE 
Content-Disposition: form-data; name="AWSAccessKeyId" 

asdfasdfFASDFSDFAADSFHHVDQ 
------WebKitFormBoundaryxh8thnHAmDhZQuXE 
Content-Disposition: form-data; name="policy" 

FsnY29udFuZ2UnLCAwLCAxMDAwMDAwMDBdLAogICAgICasdfasdfAgIFsgJ3N0YXJ0cy13aXRoJywgJyRrZXknLCAnJyBdLAogICAgICAgIFsgJ3N0YXJ0cy13aXRoJywgJyRDb250ZW50LVR5cGUnLCAnasdfJyBdLAo 
------WebKitFormBoundaryxh8thnHAmDhZQuXE 
Content-Disposition: form-data; name="name" 

Policy134722343242345 
------WebKitFormBoundaryxh8thnHAmDhZQuXE 
Content-Disposition: form-data; name="success_action_status" 

201 
------WebKitFormBoundaryxh8thnHAmDhZQuXE 
Content-Disposition: form-data; name="signature" 

basdfasdftwa/9asdfasdfx3/zasdfadsft6g= 
------WebKitFormBoundaryxh8thnHAmDhZQuXE 
Content-Disposition: form-data; name="file"; filename="blob" 
Content-Type: image/jpeg 


------WebKitFormBoundaryxh8thnHAmDhZQuXE 
Content-Disposition: form-data; name="filename" 

C:\fakepath\495845894.jpg 
------WebKitFormBoundaryxh8thnHAmDhZQuXE-- 

回答

7

找出問題,爲formdata排序很重要,您必須按照正確的順序才能正確發佈數據。

+1

從[AWS Docs](http://doc.s3.amazonaws.com/proposals/post.html#Variation_on_the_form) - 該文件應該是表單中的最後一個元素。我花了一段時間才找到這份文件。所以這裏是... – arty 2014-04-24 11:34:16

+0

我使用blueimp jQuery文件上傳並直接將裁剪圖像上傳到s3。 這工作對我來說追加一個文件屬性到'add'事件內的formData。 ''' data.formData = { ACL: '公衆閱讀', AWSAccessKeyId:res.s3Key, 鍵:IMAGE_PATH, 政策:res.s3PolicyBase64, 簽名:res.s3Signature, 文件:數據。檔案[0] }''' – looshi 2016-01-23 21:20:15