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?

The error 500 is most likely caused by $test=1.1.1.1, the rest of your code is FINE You need to add string quotes around the $test variable, such as: $test = '1.1.1.1' — Constantin
– Constantin, Commented May 16, 2020 at 11:21
The posted answer is correct. I'll also add that "$test" is the same as simply $test and the latter looks a lot cleaner. Also you're mixing up styles. array() is the old style and [] is the newer style. I suggest you stick to one or the other (and the newer style is a lot cleaner). So - $data = ["ips"=>["ip"=>$test]]; — Vilx-
– Vilx-, Commented May 16, 2020 at 11:23

Mike Foxtech · Accepted Answer · 2020-05-16 11:20:56Z

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);

answered May 16, 2020 at 11:20

Mike Foxtech

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DannyM · Accepted Answer · 2020-05-16 11:21:37Z

1

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"]];

answered May 16, 2020 at 11:21

DannyM

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1 Comment

Harsh Over a year ago

Oh shit, i am so stupid. I was focusing on the $data that i completely missed the first one.

mistapink · Accepted Answer · 2020-05-16 11:29:46Z

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.

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