我不明白如何用PHP正確創建並返回有用的錯誤信息到Web。用PHP返回有用的錯誤信息
我有一個類
class Foo {
const OK_IT_WORKED = 0;
const ERR_IT_FAILED = 1;
const ERR_IT_TIMED_OUT = 3;
public function fooItUp(){
if(itFooed)
return OK_IT_WORKED;
elseif(itFooedUp)
return ERR_IT_FAILED;
elseif(itFooedOut)
return ERR_IT_TIMED_OUT;
}
}
和使用這個類做一些有用的另一個類,然後將結果返回給用戶。我只是想知道我把所有的錯誤消息的字符串值。
class Bar {
public function doFooeyThings(stuff){
$res = $myFoo->fooItUp();
// now i need to tell the user what happened, but they don't understand error codes
if($res === Foo::OK_IT_WORKED)
return 'string result here? seems wrong';
elseif ($res === Foo::ERR_IT_FAILED)
return Foo::ERR_IT_FAILED_STRING; // seems redundant?
elseif($res === Foo:ERR_IT_TIMED_OUT)
return $res; // return number and have an "enum" in the client (js) ?
}
}