how to use variables in include files

Question

Is it possible to include files with a variable inside of include?

See this example:

$randfilescnt = rand(1,$filecount);

include('./quizes/Arrow/quiz'+ $randfilescnt +'.php');

How can I do this?

That would work if you were using the right concatenator. `+` is for javascript, `.` is for PHP. — aynber, Nov 01 '17 at 13:08
@rajdevkumar are the javascript, html and css tags relevant to the question? If not, please remove them — GreyRoofPigeon, Nov 01 '17 at 13:11

score 1 · Answer 1 · answered Nov 01 '17 at 13:09

1

Like this way :

$randfilescnt = rand(1,$filecount);
include("./quizes/Arrow/quiz".$randfilescnt.".php");

answered Nov 01 '17 at 13:09

Naushil Jain

it works thanks <3 – rajdev kumar Nov 01 '17 at 13:15
1

@rajdevkumar You are meant to accept the answer if it helped you. – SomeBruh Nov 01 '17 at 13:16
@TNU ? i did n´t understand – rajdev kumar Nov 01 '17 at 14:38

score 0 · Accepted Answer · answered Nov 01 '17 at 13:08

0

Use "." for string concat in php

 $randfilescnt = rand(1,$filecount);

    include('./quizes/Arrow/quiz'.$randfilescnt.'.php');

answered Nov 01 '17 at 13:08

Shital Marakana

2 Answers2