json, php - output string from array

Question

I am trying to decode JSON data to PHP then output it to the site. If I have the following:

{
 "name": "josh",
 "type": "human"
{

I can do this (within PHP), to display or output my type:

$file = "path";
$json = json_decode($file);
echo $json["type"]; //human

So, if I have the following:

{
 "name": "josh",
 "type": "human",
 "friends": [
 {
 "name": "ben",
 "type": "robot"
 },
 {
 "name": "tom",
 "type": "alien"
 }
 ],
 "img": "img/path"
}

How can I output what type my friend ben is?

• sjsam: You are encouraged to use jq that is best suited for such jobs. You might want to check my [ other answer ] on a similar question.

Answer 1

Use a loop like foreach and do something like the following:

//specify the name of the friend like this:
$name = "ben";
$friends = $json["friends"];
//loop through the array of friends;
foreach($friends as $friend) {
 if ($friend["name"] == $name) echo $friend["type"];
}

Answer 2

To get the decoded data in array format you would supply true as the second argument to json_decode otherwise it will use the default which is object notation. You could easily create a function to shorten the process when you need to find a specific user

$data='{
 "name": "josh",
 "type": "human",
 "friends": [
 {
 "name": "ben",
 "type": "robot"
 },
 {
 "name": "tom",
 "type": "alien"
 }
 ],
 "img": "img/path"
}';
$json=json_decode($data);
$friends=$json->friends;
foreach( $friends as $friend ){
 if( $friend->name=='ben' )echo $friend->type;
}
function finduser($obj,$name){
 foreach( $obj as $friend ){
 if( $friend->name==$name )return $friend->type;
 }
}
echo 'Tom is a '.finduser($friends,'tom');

Answer 3

try this,

$friend_name = "ben";
$json=json_decode($data);
$friends=$json->friends;
foreach( $friends as $val){
 if($friend_name == $val->name)
 {
 echo "name = ".$val->name;
 echo "type = ".$val->type;
 } 
}

DEMO