Echo variable in json/array in PHP

Question

I have made a code that works however it doesn't work when I try to echo a variable in the json/array. Can anyone help?

Code that works

$data = array("ips" => ["ip" => "1.1.1.1"]);                                                                    
$data_string = json_encode($data);  

Code that doesn't work

 $test=1.1.1.1
 $data = array("ips" => ["ip" => "$test"]);                                                                    
 $data_string = json_encode($data);  

The code just return with 500error on browser

As you can see i am trying to introduce a variable. Can anyone help?

Answer 1

You have syntax error in $test variable. You can use string type, like this:

$test = '1.1.1.1';
$data = array("ips" => ["ip" => "$test"]);                                                                    
$data_string = json_encode($data);  

Answer 2

you need to put quotes around "1.1.1.1", or php's parser doesn't interpret it as a string (which is what it's supposed to be)

DO THIS:

$test="1.1.1.1";
$data = array("ips" => ["ip" => "$test"]);                                                                    
$data_string = json_encode($data);  

NOT THIS:

$test=1.1.1.1
$data = array("ips" => ["ip" => "$test"]);                                                                    
$data_string = json_encode($data);  

Also it's bad practice to mix the shorthand for arrays with the array function, you should either do array("ips" => array("ip" => "$test")); or ["ips" => ["ip" => "$test"]];

Answer 3

While the right answers are already given: You have a syntax error there and should write

$ip = "1.1.1.1";

I would recommend that you first take a look at your server's error logs for the web server and PHP. It will clearly state why your script is failing:

[Sat May 16 13:26:20.744739 2020] [proxy_fcgi:error] [pid 1653:tid 140650782840576] [client ::1:55866] AH01071: Got error 'PHP message: PHP Parse error:  syntax error, unexpected '.1' (T_DNUMBER) in /var/www/html/stackoverflow.php on line 2\n', referer: http://localhost/

Alternatively you can use php myScript.php and run this from the console and get the same error message. Naturally, the possible use of the PHP-CLI depends a bit on the complexity of your script.

This will help yourself in the future to find those errors easier.