使用收取($_REQUEST[p])或$_REQUEST[p]-Usingisset($_REQUEST[“p”])or$_REQUEST[“p”]

6  

They have error_reporting turned down, which is nice because it means you can do things like

他们有错误报告被拒绝，这很好，因为这意味着你可以做类似的事情。

if ($_POST['whatever']) { ... }

instead of

而不是

if (isset($_POST['whatever'])) { ... }

but it also stops you from seeing other possibly pertinent errors.

但它也阻止你看到其他可能的相关错误。

this setting is found in the php.ini file under the variable error_reporting.

该设置在php中找到。在变量error_reporting下的ini文件。

More information on the ini file can be found here: http://php.net/manual/en/ini.php

更多关于ini文件的信息可以在这里找到: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.

同样，isset($_REQUEST["p"])=="拉皮条"，而在sytac正确的情况下，永远不会返回true，因为isset()返回一个布尔值。

what you want is isset($_REQUEST['p']) && $_REQUEST['p'] == 'procurar'

你想要的是isset($_REQUEST['p']) & $_REQUEST['p'] = ' '

5  

RTM: http://php.net/isset

RTM:http://php.net/isset

isset() returns a boolean TRUE/FALSE. It will NEVER return a string, so this statement

isset()返回一个布尔值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()永远不会等于代理，所以['cont']检查永远不会被评估。

0  

When the first statement is false, PHP does not bother triyng the rest of the if statement.

当第一个语句为false时，PHP就不会对if语句的其余部分产生影响。

I always use the following check on every $_REQUEST, $_POST or $_GET key:

我总是在每个$_REQUEST、$_POST或$_GET键中使用以下检查:

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.

这不会给您任何警告，即使error_reporting被设置为E_ALL。

The 'isset' checks if the $key is set and after that it checks if $key has a value.

“isset”检查$key是否设置，然后检查$key是否具有值。