PHP close tag concatenate confusion

Question

eval("\$data = $myvar('https://www.example.com/json/id='". $_GET['name'] ."'))';

This got me an error, how to concatenate it properly?

Please bear in mind that `eval()` is a dangerous function. Avoid if possible. — Raptor, Jun 17 '15 at 06:31

score 0 · Accepted Answer · edited May 23 '17 at 12:05

Eval is dangerous, but for your specific question above, the quotes were off at the end, and myvar is supposed to be a function. See below:

eval('$data = myvar("https://www.example.com/json/id='. $_GET['name'] .'");');

If you use double quotes like "$data" then $data will first be evaluated and the result or value will be eval'd instead. This is one of the risks of using eval(). If you use double quotes, then escape the $ signs like so:

eval("\$data = myvar('https://www.example.com/json/id=". $_GET['name'] ."');");

Demo: IDEOne

score 0 · Answer 2 · answered Jun 17 '15 at 06:48

0

eval("\$data = $myvar('https://www.example.com/json/id=". $_GET['name']."');");

answered Jun 17 '15 at 06:48

Ramzan Meeral

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Could you please elaborate more your answer adding a little more description about the solution you provide? – abarisone Jun 17 '15 at 06:54

PHP close tag concatenate confusion

2 Answers2