在PHP中尝试catch语句,文件不上传

Question

我理解try-catch语句的作用,但是通过阅读php.net上的文档,我无法在我自己的代码中实现一个.我需要一个真实的例子来帮助我理解.

如果上传不成功,如何将此示例转换为try catch语句？

$move = move_uploaded_file($_FILES['file']['tmp_name'], $_SERVER['DOCUMENT_ROOT']."/uploads/".$_FILES['file']['name']);

if (!$move) {
    die ('File didn't upload');
} else {            
    //opens the uploaded file for extraction
    echo "Upload Complete!";
}

这可能不是一个很好的例子,但任何帮助将不胜感激.

Answer 1

你可以这样做.

try {
    //throw exception if can't move the file
    if (!move_uploaded_file( ... )) {
        throw new Exception('Could not move file');
    }

    //do some more things with the file which may also throw an exception
    //...

    //ok if got here
    echo "Upload Complete!";
} catch (Exception $e) {
    die ('File did not upload: ' . $e->getMessage());
}

对于上面的例子,这有点无意义,但你应该明白这一点.请注意,您可以从任何地方抛出异常(例如,在您使用try {}调用的函数/方法中),它们将向上传播.

Answer 2

好吧,如果你想使用例外,你可以这样做:

function handleUpload() {


    $move = move_uploaded_file($_FILES['file']['tmp_name'], $_SERVER['DOCUMENT_ROOT']."/uploads/".$_FILES['file']['name']);

    if (!$move) {
       throw new Exception('File Didnt Upload');
    }

}

try {
   handleUpload();
   echo "File Uploaded Successfully";
} catch(Exception $ex) {
   die($ex->getMessage);
}

我知道这看起来很臃肿 - 但你可以从调用堆栈的任何地方调用该方法,并在任何时候捕获异常.

Answer 3

try-catch语句用于处理异常.我不相信函数move_uploaded_files可以抛出和异常,因此我认为你编写的代码是正确的.通话结束后,查看返回代码.如果它是假的,则结束处理,否则您报告成功.