PHP：fopen错误处理 [英] PHP: fopen error handling

问题描述

我使用

$fp = fopen('uploads/Team/img/'.$team_id.'.png', "rb");
$str = stream_get_contents($fp);
fclose($fp);

然后该方法将其作为图像返回。但是当fopen（）失败时，因为文件不存在，它会引发错误：

and then the method gives it back as image. But when fopen() fails, because the file did not exists, it throws an error:

[{"message":"Warning: fopen(uploads\/Team\/img\/1.png): failed to open stream: No such file or directory in C:\...

显然，这是回到json。

This is coming back as json, obviously.

现在的问题是：捕获错误并阻止该方法直接将该错误引用到客户端？

The Question is now: How can i catch the error and prevent the method from throwing this error directly to the client?

推荐答案

您应该先测试一个文件的存在通过file_exists（）。

You should first test the existence of a file by file_exists().

    try
    {
      $fileName = 'uploads/Team/img/'.$team_id.'.png';

      if ( !file_exists($fileName) ) {
        throw new Exception('File not found.');
      }

      $fp = fopen($fileName, "rb");
      if ( !$fp ) {
        throw new Exception('File open failed.');
      }  
      $str = stream_get_contents($fp);
      fclose($fp);

      // send success JSON

    } catch ( Exception $e ) {
      // send error message if you can
    } 

或简单解决方案，无例外：

or simple solution without exceptions:

    $fileName = 'uploads/Team/img/'.$team_id.'.png';
    if ( file_exists($fileName) && ($fp = fopen($fileName, "rb"))!==false ) {

      $str = stream_get_contents($fp);
      fclose($fp);

      // send success JSON    
    }
    else
    {
      // send error message if you can  
    }

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