Question

我想在我的代码的一部分中使用异常来处理错误，但如果代码失败，我希望脚本继续。我想记录错误。有人可以帮我解决这个问题吗？

try{
    if($id == 4)
    {
      echo'test';
    }
}
catch(Exception $e){
    echo $e->getMessage();
}

echo'Hello, you should see me...'; <------ I never see this.. No errors, just a trace.

Answer 1

你必须抓住例外：

// some code

try {
    // some code, that might throw an exception
    // Note that, when the exception is thrown, the code that's after what
    // threw it, until the end of this "try" block, will not be executed
} catch (Exception $e) {
    // deal with the exception
    // code that will be executed only when an exception is thrown
    echo $e->getMessage(); // for instance
}

// some code, that will always be executed

以下是您应该阅读的一些内容：

Answer 2

在调用的代码中可能抛出Exception的代码

try { 
    // code that may break/throw an exception
    echo 'Foo';
    throw new Exception('Nothing in this try block beyond this line');
    echo 'I am never executed';
    throw new CustomException('Neither am I');
} catch(CustomException $e) {
    // continue here when any CustomException in try block occurs
    echo $e->getMessage();
} catch(Exception $e) { 
    // continue here when any other Exception in try block occurs
    echo $e->getMessage();
}

// script continues here
echo 'done';

输出将是（为便于阅读添加换行符）：

'Foo'                                         // echoed in try block
'Nothing in this try block beyond this line'  // echoed in Exception catch block
'done'                                        // echoed after try/catch block

Try / Catch Blocks也可以嵌套。请参阅上面链接的PHP手册页中的示例2：

try{
    try {
        throw new Exception('Foo');
        echo 'not getting here';
    } catch (Exception $e) {
        echo $e->getMessage();
    }
    echo 'bar';
} catch (Exception $e) {
    echo $e->getMessage();
}
echo 'done';

'Foo'  // echoed in inner catch block
'bar'  // echoed after inner try/catch block
'done' // echoed after outer try/catch block

DevZone的进一步阅读：

如何在没有脚本死亡的情况下使用异常？

2 个答案: