Can't display JSON Data in PHP - php

Question

I'm trying to use JSON in PHP and I got stucked. I'm new in this, please, if you can help me with that.
I want to display content from "list" from the code below (For example list, name and/or frags).
{
"status": true,
"hostname": "{IP}",
"port": {PORT},
"queryPort": {PORT},
"name": "Cod2 Server",
"map": "mp_genesisarc",
"secured": true,
"password_protected": false,
"version": "1.3",
"protocol": "udp",
"players": {
"online": 1,
"max": "28",
"list": [
{
"frags": "1",
"ping": "54",
"name": "Mr Anderson"
}
]
},
"cached": false
}
This is the code I use (Doesn't work):
<?php
$serverip = "localhost";
$info = json_decode( file_get_contents( 'https://use.gameapis.net/cod2/query/info/000.000.00.0:0000'.$serverip ), true );
if(!$info['status']) {
echo 'Offline';
} else {
echo $info['players']['list']['name'];
}
?>
Thanks in advance,
Jim.

Answer 1

$info['players']['list'][0]['name']
This will get you desired element. Try outputting $info next time to find this error yourself.

Answer 2

Try using a foreach loop. It's cleaner when iterating an array.
<?php
$serverip = "localhost";
$info = json_decode( file_get_contents( 'https://use.gameapis.net/cod2/query/info/000.000.00.0:0000'.$serverip ), true );
if(!$info['status']) {
echo 'Offline';
}
else {
foreach ($info['players']['list'] as $player) {
echo $player['name'];
}
}
?>

Answer 3

list will you number of online users records so there may be single or multiple users so you have to go through the loop and can display all the records available in the list. Add for loop to your code instead of displaying single record with index.
if(count($info['players']['list']) > 0){
for($cnt = 0; $cnt < count($info['players']['list']); $cnt++){
echo $info['players']['list'][$cnt]['name'];
}
} else {
echo "Write your message";
}

Answer 4

Took the following array (I had to change few things) and go here . Paste the array at the left side, then press the right arrow and see at the right side the analysis of your array. Open the tree view and you will realize that there is a 0 index in the list. So you can not access what you want by doing echo $info['players']['list']['name']; but you need to apply the index 0 of the list echo $info['players']['list'][0]['name'];
[{
"status": true,
"hostname": "{IP}",
"port": 555,
"queryPort": 555,
"name": "Cod2 Server",
"map": "mp_genesisarc",
"secured": true,
"password_protected": false,
"version": "1.3",
"protocol": "udp",
"players": {
"online": 1,
"max": "28",
"list": [
{
"frags": "1",
"ping": "54",
"name": "Mr Anderson"
}
]
},
"cached": false
}]

Answer 5

If you paste your json string into https://jsonlint.com/ and click Validate JSON, you will see that you have a couple of syntax errors; namely the values for port and queryPort.
When they are corrected, you should have something like this:
{
"status": true,
"hostname": "{IP}",
"port": "{PORT}",
"queryPort": "{PORT}",
"name": "Cod2 Server",
"map": "mp_genesisarc",
"secured": true,
"password_protected": false,
"version": "1.3",
"protocol": "udp",
"players": {
"online": 1,
"max": "28",
"list": [{
"frags": "1",
"ping": "54",
"name": "Mr Anderson"
}]
},
"cached": false
}
Now that the json is fixed, you can convert it to an array for simple processing.
Code: (Demo)
$serverip="localhost";
$array=json_decode(file_get_contents( 'https://use.gameapis.net/cod2/query/info/000.000.00.0:0000'.$serverip),true);
if(!isset($array['status']) || $array['status']===false){
echo 'Offline';
}else{
foreach($array['players']['list'] as $players){
echo "<div>Name: {$players['name']}, Frags: {$players['frags']}, Ping: {$players['ping']}</div>\n";
}
}
foreach() is a better/cleaner practice than resorting to for() which will unfortunately require count(), and a "counter" variable, and incrementation.