PHP Random Variable

Question

Using an answer found here. I am trying to randomly select a URL found for a PHP variable called $url.

Here is the current code:

$url1 = "https://www.domain1.com/tds/parser.php?station=".$_GET['id'];
$url2 = "https://www.domain2.com/tds/parser.php?station=".$_GET['id'];

$vals = array($url1,$url2);

$url = array($vals[array_rand($vals, 1)]);

echo $url;

My echo statement returns just the words 'array' instead of the actual url randomly selected?

You didn't select a URL, you just shuffled the array. Now you can use echo $url[0] — aynber
– aynber, Commented Dec 26, 2018 at 19:36
$url = array($vals[array_rand($vals, 1)]); is confusing and probably unnecessary. Just shuffle($vals); and echo $vals[0]; — MonkeyZeus
– MonkeyZeus, Commented Dec 26, 2018 at 19:37
your shuffle is useless because at this point, your array.length === 1 — Yanis-git
– Yanis-git, Commented Dec 26, 2018 at 19:37
I don't want to display the url with an echo, I want to make that variable available for later in the program after it's randomly shuffled. — Rocco The Taco
– Rocco The Taco, Commented Dec 26, 2018 at 19:38

ash__939 · Accepted Answer · 2018-12-26 20:55:10Z

1

The problem is with this line:
$url = array($vals[array_rand($vals, 1)]);

This will assign an array to the variable $url

Change it to $url = $vals[array_rand($vals, 1)];

<?php

$vals = array(
    "https://www.domain1.com/tds/parser.php?station=".$_GET['id'],
    "https://www.domain2.com/tds/parser.php?station=".$_GET['id']
);

$url = $vals[array_rand($vals, 1)];

echo $url;

answered Dec 26, 2018 at 20:55

ash__939

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