Question

如何在构造函数中使用错误控制运算符（@）？

\DateTime的构造函数将抛出异常并在传递错误参数时发出警告，但我宁愿能够catch异常并处理它，并且不会收到警告。出于这个原因，我想沉默警告，例如使用@。

如果我发现异常，则仍会发出警告。我可以尝试使用@运算符（虽然我不知道它是否也可以阻止异常被引发）。但我无法获得此用例的语法。

正在测试的PHP版本是7.0.22。

❯ php -a
Interactive shell

php > new DateTime("abc");
PHP Warning:  Uncaught Exception: DateTime::__construct(): Failed to parse time string (abc) at position 0 (a): The timezone could not be found in the database in php shell code:1
Stack trace:
#0 php shell code(1): DateTime->__construct('abc')
#1 {main}
  thrown in php shell code on line 1

Warning: Uncaught Exception: DateTime::__construct(): Failed to parse time string (abc) at position 0 (a): The timezone could not be found in the database in php shell code:1
Stack trace:
#0 php shell code(1): DateTime->__construct('abc')
#1 {main}
  thrown in php shell code on line 1

php > $a = @new DateTime("abc");
PHP Warning:  Uncaught Exception: DateTime::__construct(): Failed to parse time string (abc) at position 0 (a): The timezone could not be found in the database in php shell code:1
Stack trace:
#0 php shell code(1): DateTime->__construct('abc')
#1 {main}
  thrown in php shell code on line 1

Warning: Uncaught Exception: DateTime::__construct(): Failed to parse time string (abc) at position 0 (a): The timezone could not be found in the database in php shell code:1
Stack trace:
#0 php shell code(1): DateTime->__construct('abc')
#1 {main}
  thrown in php shell code on line 1

php > $a = @(new DateTime("abc"));
PHP Warning:  Uncaught Exception: DateTime::__construct(): Failed to parse time string (abc) at position 0 (a): The timezone could not be found in the database in php shell code:1
Stack trace:
#0 php shell code(1): DateTime->__construct('abc')
#1 {main}
  thrown in php shell code on line 1

Warning: Uncaught Exception: DateTime::__construct(): Failed to parse time string (abc) at position 0 (a): The timezone could not be found in the database in php shell code:1
Stack trace:
#0 php shell code(1): DateTime->__construct('abc')
#1 {main}
  thrown in php shell code on line 1
php > 

php > $a = new @DateTime("abc");
PHP Parse error:  syntax error, unexpected '@' in php shell code on line 1

Parse error: syntax error, unexpected '@' in php shell code on line 1

Answer 1

为什么不使用try catch？

try {
    $dt = new DateTime("abc");
} catch(Exception $e) {
    // do wathever you want
}

Answer 2

你可以做的是使用set_error_handler的自定义处理函数，这样你就可以捕获＆＃34;你想要的警告/错误。

这是一个我用来测试正则表达式是否有效的例子，我只是使用一个空函数，因为我不需要对警告做任何事情：

function filterRegularExpression($str) {
    set_error_handler(function() {}, E_WARNING);
    $isRegularExpression = preg_match($str, '') !== false;
    restore_error_handler();
    if($isRegularExpression){
        return $str;
    }
    return false;
}

Answer 3

您误解了产生警告的内容。警告由全局错误处理程序生成，因为您没有捕获异常并且它冒泡到顶部以终止程序（或者，在交互式CLI上生成警告）。线索是警告的前缀是“Uncaught Exception”。如果捕获异常，则不会产生警告。 DateTime构造函数不会产生自己的警告。

在PHP中使用带有构造函数的错误控制操作符

3 个答案: