How to add string to a PHP variable? - php

Question

I'm not sure it's even the right way to define this question.
I have string that may be exist, and may not. It happens to be a number: $number
If $number doesn't exist, then I want to use the PHP variable $url.
But if $number does exist, then I want to use the PHP variable which is named $url+the number, i.e, $url2 if $number=2
So I tried this code, but it doesn't work:
$number = "2"; //(Can be either missing, or equal to 1, 2, or 3)
$url = "www.0.com"; // Fallback
$url1 = "www.1.com";
$url2 = "www.2.com";
$url3 = "www.3.com";
$result = $url.=$number ;
// If $number=1, I want $result to be : www.1.com
// If $number=2, I want $result to be : www.2.com
// If $number=3, I want $result to be : www.3.com
// If $number IS NOT SET, I want $result to be : www.0.com
// Now do something with $result
Perhaps there's a completely better way to achieve what I want (will be happy to see example), but anyway I'm curious as well to understand how to achieve it my way.

Answer 1

Okay, so you're talking about a variable variable.
You should define the name of the variable you need to use in a string, and then pass that to a variable variable using $$ syntax:
if( isset($number) && is_numeric($number) )
{
$name = 'url'.$number;
$result = $$name;
}
else
{
$result = $url;
}
That having been said, you may be better off using an array for this:
$urls = [ 'www.0.com', 'www.1.com', 'www.2.com', 'www.3.com' ];
$result = (!isset($number)) ? $urls[0] : $urls[ intval($number) ];

Answer 2

You can use ternary with in_array and empty.
$number = "2"; //(Can be either missing, or equal to 1, 2, or 3)
$url = "www.0.com"; // Fallback
$url1 = "www.1.com";
$url2 = "www.2.com";
$url3 = "www.3.com";
$result = (!empty($number) && in_array($number, array(1,2,3))) ? ${'url' . $number} : $url;
echo $result;
Demo: https://eval.in/821737

Answer 3

In php you can have things like dynamic variable names:
$variableName = "url".$number;
$result = $$variableName;
However, you should make sure, that $variableName refers to an existing variable:
$result = "www.fallbackURL.com";
if(isset($$variableName)) $result = $$variableName;

Answer 4

Or Try this code:
$number = 5;
$url[0] = "www.0.com"; // Fallback
$url[1] = "www.1.com";
$url[2] = "www.2.com";
$url[3] = "www.3.com";
if (!isset($number)) $number=0;
if (!isset($url[$number])) $number=0;
$result = $url[$number];

Answer 5

If you add $ front of string, it define variable, so you can use following code:
<?php
$number = "2"; //(Can be either missing, or equal to 1/2/3)
$url = "www.0.com"; // Fallback
$url1 = "www.1.com";
$url2 = "www.2.com";
$url3 = "www.3.com";
if(isset($number) && is_numeric($number) && $number <= 3) {
$variable_name = 'url' . $number; //string like url2
} else {
$variable_name = 'url';
}
$result = $$variable_name ; //define $url2 from url2 string
echo $result;
// Now do something with $result
Example for define variable with string variable:
$string = 'hello';
$$string = 'new variable'; //define $hello variable
echo $hello; //Output: "new variable"

Answer 6

if the url need just a number, you can do this easy way
($number)?$number:0;
$url = "www.".$number.".com";
if there are specific real url, you can use array
$array[0] = "www.google.com";
$array[1] = "www.facebook.com";
($number)?$number:0;
url = $array[$number];

Answer 7

Updated code:
$number = "2";
if(isset($number)){
$res = "url".$number;
$result=$$res;
}else{
$result=$url;
}
echo $result;