更正 PHP 代码以检查变量是否存在

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时间:2020-08-25 22:41:22  来源:igfitidea点击:

Correct PHP code to check if a variable exists

phpjsonwebsocket

提问by JJJollyjim

This code is part of a websocket server:

此代码是 websocket 服务器的一部分:

$msgArray = json_decode($msg);
if ($msgArray->sciID) {
  echo "Has sciID";
}

It will either be receiving a json string like {"sciID":67812343}or a completely different json string with no sciID such as {"something":"else"}.

它将接收类似 json 字符串{"sciID":67812343}或完全不同的没有 sciID 的 json 字符串,例如{"something":"else"}.

When the server receives the later, it echos out: Notice: Undefined property: stdClass::$sciID in /path/to/file.php on line 10

当服务器接收到后者时,它会回显: Notice: Undefined property: stdClass::$sciID in /path/to/file.php on line 10

What is the correct code to check if $msgArray->sciIDexists?

检查是否$msgArray->sciID存在的正确代码是什么?

回答by Ross

Use issetas a general purpose check (you could also use property_existssince you're dealing with an object):

使用isset作为通用支票(你也可以使用property_exists,因为你正在处理的对象):

if (isset($msgArray->sciID)) {
    echo "Has sciID";
}

回答by Oliver Charlesworth

回答by Petr Hurtak

In case isset()or property_exists()doesn't work, we can use array_key_exists()

万一isset()property_exists()不起作用,我们可以使用array_key_exists()

if (array_key_exists("key", $array)) {
    echo "Has key";
}

回答by Joshwaa

I've always done isset()but I've recently changed to !empty()and emptybecause one of my friends suggested it.

我一直这样做,isset()但我最近改为!empty()并且empty因为我的一位朋友建议这样做。

回答by Sophivorus

The isset()function checks if a variable is set and is not null, which is really stupid, given its name. If you only want to know if a variable is set, use the following function:

isset()函数检查变量是否已设置并且不为 null,鉴于其名称,这真的很愚蠢。如果您只想知道是否设置了变量,请使用以下函数:

function is_set( & $variable ) {
    if ( isset( $variable ) and ! is_null( $variable ) )
        return true;
    else
        return false;
}

This function will return true exactly when the variable is not set, and false otherwise. The '&' next to the argument is important, because it tells PHP to pass a referenceto the variable, and not the valueof the variable, thus avoiding a "Notice:Undefined variable" when the variable is not set.

当变量未设置时,此函数将返回 true,否则返回 false。参数旁边的 '&' 很重要,因为它告诉 PHP 传递对变量的引用,而不是变量的,从而避免在未设置变量时出现“注意:未定义变量”。