从PHP写入映像文件时出错

问题描述

我正在尝试从blob写入图像文件。

I am attempting to write to an image file from a blob.

 if($_POST['logoFilename'] != 'undefined'){
  $logoFile = fopen($_POST['logoFilename'], 'w') or die ("Cannot create ".$_POST['logoFilename']);

  fwrite($logoFile, $_POST['logoImage']);

  fclose($logoFile);
}

在上一段代码中， $ _ POST [' logoImage'] 是一个BLOB。该文件已正确写入根目录，但无法打开该文件。在ubuntu 11.04中，我收到以下错误：

In the previous code snippet, $_POST['logoImage'] is a BLOB. The file is correctly written to the root directory, however the file cannot be opened. In ubuntu 11.04 I receive the following error:

Error interpreting JPEG image file (Not a JPEG file: starts with 0x64 0x61).

如果我创建一个img并设置其src = blob

The BLOB does correctly display if I create an img and set its src=blob

下面是BLOB的第一个片段：

Included below is the first snippet of the BLOB:

推荐答案

你的实际上是：

data:[<MIME-type>][;charset=<encoding>][;base64],<data>

由于你只需要解码数据部分，你必须这样做

Since you only want the decoded data part, you have to do

file_put_contents(
    'image.jpg',
    base64_decode(
        str_replace('data:image/jpeg;base64,', '', $blob)
    )
);

但是由于PHP本身支持data：// streams，你也可以这样做（感谢@NikiC）

But since PHP natively supports data:// streams, you can also do (thanks @NikiC)

file_put_contents('image.jpg', file_get_contents($blob));

如果上述方法不起作用，您可以尝试使用GDlib：

If the above doesnt work, you can try with GDlib:

imagejpg(
    imagecreatefromstring(
        base64_decode(
            str_replace('data:image/jpeg;base64,', '', $blob)
        )
    ),
    'image.jpg'
);

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