json, php - output string from array - php

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?

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