PHP：fopen 错误处理

PHP: fopen error handling

提问by Fabian

I do fetch a file with

我确实获取了一个文件

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

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

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，它会抛出一个错误：

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

This is coming back as json, obviously.

显然，这将作为 json 返回。

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

现在的问题是：如何捕获错误并防止该方法直接将此错误抛出给客户端？

回答by Cendak

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

您应该首先通过 file_exists() 测试文件是否存在。

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
} 

or simple solution without exceptions:

或无一例外的简单解决方案：

$fileName = 'uploads/Team/img/'.$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  
}

回答by Refazul Refat

You can use the file_exists()function before calling fopen().

您可以在调用fopen()之前使用file_exists()函数。

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

回答by Sat93

Generically- This is probably the best way to do file-io in php (as mentioned by @Cendak here)

一般- 这可能是在 php 中执行 file-io 的最佳方法（如 @Cendak在这里提到的）

$fileName = 'uploads/Team/img/'.$team_id.'.png';
if ( file_exists($fileName) && ($fp = fopen($fileName, "rb"))!==false ){
    $str = stream_get_contents($fp);
    fclose($fp);
    // send success JSON    
}else{
    // send an error message if you can  
}

But it does not work with PHP 7.3, these modifications do,

但它不适用于 PHP 7.3，这些修改可以，

if(file_exists($filename) && ($fp = fopen($filename,"r") !== false)){
        $fp = fopen($filename,"r");
        $filedata = fread($fp,filesize($filename));
        fclose($fp);
}else{
        $filedata = "default-string";
}

回答by yyg

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

the error is clear: you've put the wrong directory, you can try what you whant but it'll not work. you can make it work with this:

错误很明显：您放置了错误的目录，您可以尝试您想要的，但它不起作用。你可以让它工作：

1. take your file and put it in the same folder of your php file (you'll be able to move it after don't worry, it's about your error) or on a folder "higher" of your script (just not outside of your www folder)
2. change the fopen to ('./$team_id.'png',"rb");
3. rerun your script file

1. 把你的文件放在你的 php 文件的同一个文件夹中（不用担心，你可以移动它，这是关于你的错误）或你的脚本的“更高”文件夹（只是不在你的文件夹之外） www 文件夹）
2. 将 fopen 更改为 ('./$team_id.'png',"rb");
3. 重新运行你的脚本文件

don't forget this : you can't access a file that is'nt in your "www" folder (he doesn't found your file because he give you her name: the name come from the $team_id variable)

不要忘记这一点：您无法访问不在“www”文件夹中的文件（他没有找到您的文件，因为他给了您她的名字：名称来自 $team_id 变量）