Ajax：發送沒有實際文件的文件信息

我在寫一個web應用程序來替換C++程序。該應用程序的目標是從瀏覽器添加或拍照，然後將其上傳到服務器。Ajax：發送沒有實際文件的文件信息

上傳文件有兩個主要步驟，檢查服務器上是否已存在文件，如果沒有，則上傳文件。

我想知道如何在Ajax中發送第一個主要步驟（發送文件（PHP腳本使用$ _FILES，我無法更改）而沒有實際的文件內容）。

C++代碼看起來是這樣的：

*dataBuffer_ += QByteArray("--") + QByteArray(MULTIPART_BOUNDARY) + QByteArray("\r\n"); 
*dataBuffer_ += QByteArray("content-disposition: form-data; name=\"expected_size\"\r\n\r\n"); 
*dataBuffer_ += fileSize.toAscii() + QByteArray("\r\n"); 

//original size 
//Original size of the file before compression and encoding 
*dataBuffer_ += QByteArray("--") + QByteArray(MULTIPART_BOUNDARY) + QByteArray("\r\n"); 
*dataBuffer_ += QByteArray("content-disposition: form-data; name=\"original_size\"\r\n\r\n"); 
*dataBuffer_ += QString::number(file.size()).toAscii() + QByteArray("\r\n"); 

//tell upload.php that we have a file to upload 
*dataBuffer_ += QByteArray("--") + QByteArray(MULTIPART_BOUNDARY) + QByteArray("\r\n"); 
*dataBuffer_ += QByteArray("content-disposition: form-data; name=\"mode\"\r\n\r\n"); 
*dataBuffer_ += QByteArray("check_file\r\n"); 

//the binary 
*dataBuffer_ += QByteArray("--") + QByteArray(MULTIPART_BOUNDARY) + QByteArray("\r\n"); 
*dataBuffer_ += QByteArray("content-disposition: form-data; name=\"file\"; filename=\"") + file.fileName().toAscii() + QByteArray("\"\r\n"); 
*dataBuffer_ += QByteArray("content-type: application/zip\r\n\r\n"); 
*dataBuffer_ += fileBuffer; 
*dataBuffer_ += QByteArray("\r\n"); 

*dataBuffer_ += QByteArray("--") + QByteArray(MULTIPART_BOUNDARY) + QByteArray("--");

我願做這樣的事情：

var form = new FormData(); 
form.append("expected_size", expected_size); 
form.append("original_size", original_size); 
form.append("mode", "check_file"); 
form.append("file", fileInput.files[0]); 

var xhr = new XMLHttpRequest(); 
xhr.open('POST', action, true); 
xhr.send(formData);

該腳本將用實際文件的文件信息，而我只是想發送文件信息。任何人都知道我該怎麼做？

下面是PHP腳本（我不能修改）：

//check if the file was already uploaded to the server (for resuming an upload) 
//returns file:exist or file:not_exist 
//the post is the same as when transmitting a file, except the data block 
if ('check_file' == $_POST['mode']) 
{ 
    echo "CHECK FILE\n"; 
    if (!file_exists($_SESSION['localdir'])) 
    { 
    echo "upload:fail\n"; 
    echo "message:Unable to copy data on server\n"; 
    exit(); 
    } 

    if (empty($_FILES)) 
    { 
    echo "upload:fail\n"; 
    echo "message:No files were uploaded to the server\n"; 
    exit(); 
    } 

    $file = $_FILES['file']; 
    $path = $_SESSION['localdir'] . "/" . $file['name']; 

    if (file_exists($path)) 
    { 
     echo $file['name'] . " already exists on the server \n"; 
     echo "Server file size: " . filesize($path) . "\n"; 
     echo "Uploaded file size: " . $_POST['original_size'] . " \n"; 

     if (filesize($path) == $_POST['original_size']) 
     { 
      echo "file:exist\n"; 

      //add the file to the global list if it is not already there 
      if (!array_search($file['name'], $_SESSION['filenames'], true)) //true=strict 
       array_push($_SESSION['filenames'], $file['name']); 
     } 
     else 
      echo "file:not_exist\n"; 
    } 


}

來源

2013-04-11 Philippe Maes

如果您只關心文件名和文件大小，可以將'files [0] .name'和'files [0] .size'發送到服務器。 – Passerby 2013-04-11 10:17:44

你的意思是像form.append（「file」，fileInput.files [0] .name）？我不知道這是可能的！謝謝，我會試試:)。 – 2013-04-11 10:23:43

請注意，'files [0] .name'是一個**字符串**，'files [0] .size'是一個**整數**。你將不得不修改你的服務器端代碼來滿足這個變化（因爲沒有'$ _FILES'可用）。 – Passerby 2013-04-11 10:27:19

我通過手動設置我的頭文件來完成它（如C++上面的代碼），我的AJAX：

var boundary = generateBoundary(); 

var databuffer = "\r\n"; 
// set original_size 
databuffer += "--" + boundary + "\r\n"; 
databuffer += "content-disposition: form-data; name=\"original_size\"\r\n\r\n"; 
databuffer += allFiles[index].size + "\r\n"; 
// set mode 
databuffer += "--" + boundary + "\r\n"; 
databuffer += "content-disposition: form-data; name=\"mode\"\r\n\r\n"; 
databuffer += "check_file\r\n"; 
// set the file 
databuffer += "--" + boundary + "\r\n"; 
databuffer += "content-disposition: form-data; name=\"file\"; filename=\"" + allFiles[index].name + "\"\r\n"; 
databuffer += "content-type: application/zip\r\n\r\n"; 
databuffer += ""; // this is de file content => empty 
databuffer += "\r\n"; 
// end 
databuffer += "--" + boundary + "--";

這將發送文件與所需的信息沒有實際的文件內容。我希望這可以幫助其他人。

來源

2013-04-29 17:51:01

Ajax：發送沒有實際文件的文件信息

回答

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