Retrieve data from JSON based on the object index in PHP [duplicate] - php

Question

This question already has an answer here:
How to extract and access data from JSON with PHP?
(1 answer)
Closed 5 years ago.
I have a $abcd variable and the following is the output:
echo $abcd;
//Output:
{
"NID": 2,
"STS": "3",
"Options": {
"Model": "model value",
"Location": "location value",
"Price": "price value",
"Name": "Value"
}
}
In the "Options" I have 3 names and values of each. The names are not fixed and could be anythings and the number of objects in Options could be any from 0 to 100.
I'd like to know if there is any way (JSON format preferred) that I can assign the Names and their related values to two other variables.
$varName[0]=Model
$varValue[0]=model value
$varName[1]=Location
$varValue[1]=location value
$varName[2]=Price
$varValue[2]=price value

Answer 1

i'm not sure what is th logic behind this,
but you may accomplish this by using mix of array_key & array_values like following:
$abcd = '{
"NID": 2,
"STS": "3",
"Options": {
"Model": "model value",
"Location": "location value",
"Price": "price value",
"Name": "Value"
}
}';
$data = json_decode($abcd, true);
$keys = array_keys($data['Options']);
$values = array_values($data['Options']);
echo $keys[0] . " - " . $values[0]; // Model - model value
echo $keys[1] . " - " . $values[1]; // Location - location value
// ..... and so on.
live example: https://3v4l.org/GggRd

Answer 2

$data = json_decode($abcd, true);
foreach($data['Options'] as $index => $value) {
//do whatever you want
}
//or access them directly
$data['Options']['Model']

Answer 3

You dont need to assign them to anything else, just json_decode() the JSONString and they become a PHP Object you can use as is
$obj = json_decode($abcd);
echo $obj->Options->Model;
echo $obj->Options->Location;
echo $obj->Options->Price;
RE your comment#
I said that the names are varies and Model, Location and Price is just an example
then you could do
foreach ($obj->Options as $prop => $val) {
sprintf( '%s = %s', $prop, $val );
}