如何将$符号添加到字符串以使其成为变量?
Eg:
$consumer = array()
$industrial = array()//These 2 are in a separate include file.
$var = $_GET['val'] // value here is 'consumer'
function ('$'.$var,$bar) //I'm trying to make consumer -> $consumer
如何将$符号添加到字符串以使其成为变量?
Eg:
$consumer = array()
$industrial = array()//These 2 are in a separate include file.
$var = $_GET['val'] // value here is 'consumer'
function ('$'.$var,$bar) //I'm trying to make consumer -> $consumer
$consumer = array()
$industrial = array()//These 2 are in a separate include file.
$var = $_GET['val'] // value here is 'consumer'
function ($$var,$bar) //I'm trying to make consumer -> $consumer
不要忘记检查$ _GET ['val']的值是您(程序员)期望的值,而不是别的。
$ var =(in_array($ whitelist,$ _GET ['val'])?$ _GET ['val']:$ fallback; –
为什么不干脆:
if ($_GET['val'] == 'customer') {
function($bar);
}
不是最好的方式,以达到该值,但PHP支持:$$ VAR :)
只需执行以下操作:
$ var ='myString';
$ {$ var} ='output';
echo $ myString;
好吧,这里的白名单,你应该肯定包括。
$whitelist = array('customer', 'consumer');
$fallback = $whitelist[0];
$var = in_array($whitelist, $_GET['val'] ? $_GET['val'] : $fallback;
你错过了每一个分号。 – tekknolagi
请确保您对此应用白名单,否则用户可以自由访问您的内部PHP变量,从而导致可能的安全漏洞。 –
请不要开始分号讨论。 –