php 使用 isset($_REQUEST["p"]) 或 $_REQUEST["p"]
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Using isset($_REQUEST["p"]) or $_REQUEST["p"]
提问by calexandru
Sometime ago, I was in a internship, and I was working as a junior web-developer. While working and learning, I noticed that when changing pages, instead of using isset($_POST/GET/REQUEST["var"])they just used $_POST/GET/REQUEST["var"].
前段时间，我在实习，当时我是一名初级 Web 开发人员。在工作和学习时，我注意到在更改页面时，isset($_POST/GET/REQUEST["var"])他们只是使用$_POST/GET/REQUEST["var"].
So, later I came home, and tried the same thing. What happens ? Every-time I come across a if()to verify that, I have to use isset(), otherwhise it gives me an error. But notice one thing, my url is this:
所以，后来我回到家，并尝试了同样的事情。发生什么了 ？每次我遇到 aif()来验证时，我都必须使用isset()，否则它会给我一个错误。但请注意一件事，我的网址是这样的：
?p=sub_artigo&id=2
So, when I do the if()condition:
所以，当我做if()条件时：
if(isset($_REQUEST["p"])=="procurar" && $_REQUEST['cont']){
It doesn't show errors, but if I take of the isset(), it gives the usual error that I see in the forums and here.
它不显示错误，但如果我选择isset()，它会给出我在论坛和此处看到的常见错误。
So my question is, why doesn't show the error for the second variable ?
所以我的问题是，为什么不显示第二个变量的错误？
Note: p->string;id->int
注意：p->string;id->int
回答by castis
They have error_reportingturned down, which is nice because it means you can do things like
他们error_reporting拒绝了，这很好，因为这意味着你可以做这样的事情
if ($_POST['whatever']) { ... }
instead of
代替
if (isset($_POST['whatever'])) { ... }
but it also stops you from seeing other possibly pertinent errors.
但它也会阻止您看到其他可能相关的错误。
this setting is found in the php.inifile under the variable error_reporting.
此设置位于php.ini变量下的文件中error_reporting。
More information on the ini file can be found here: http://php.net/manual/en/ini.php
可以在此处找到有关 ini 文件的更多信息：http: //php.net/manual/en/ini.php
also, isset($_REQUEST["p"])=="procurar"while sytactically correct, is never going to return true, because isset()returns a boolean value.
此外，isset($_REQUEST["p"])=="procurar"虽然在语法上是正确的，但永远不会返回 true，因为isset()返回的是一个布尔值。
what you want is isset($_REQUEST['p']) && $_REQUEST['p'] == 'procurar'
你想要的是 isset($_REQUEST['p']) && $_REQUEST['p'] == 'procurar'
回答by Marc B
RTM: http://php.net/isset
RTM：http: //php.net/isset
isset() returns a boolean TRUE/FALSE. It will NEVERreturn a string, so this statement
isset() 返回布尔值 TRUE/FALSE。它永远不会返回一个字符串，所以这个语句
if(isset($_REQUEST["p"])=="procurar" && $_REQUEST['cont']){
can NEVER succeed, because isset() will never EVER be equal to procurar, so the ['cont']check will never be evaluated.
永远不会成功，因为 isset() 永远不会等于procurar，因此['cont']永远不会评估检查。
回答by IT-Smart
When the first statement is false, PHP does not bother triyng the rest of the if statement.
当第一条语句为假时，PHP 不会费心去尝试 if 语句的其余部分。
I always use the following check on every $_REQUEST, $_POST or $_GET key:
我总是对每个 $_REQUEST、$_POST 或 $_GET 键使用以下检查：
function ifSet($key) {
return (isset($_REQUEST[$key]) && $_REQUEST[$key])?$_REQUEST[$key]:'';
}
This will never give you any warnings, even if error_reporting is set to E_ALL.
这永远不会给你任何警告，即使 error_reporting 设置为 E_ALL。
The 'isset' checks if the $key is set and after that it checks if $key has a value.
'isset' 检查是否设置了 $key，然后检查 $key 是否有值。
