Adding all state in dropdown from JSON file based on country selection - php

Question

I have list of countries and their states in JSON file, below is how it looks like
"NO": {
"name": "Norway",
"states": {
"VA": {
"name": "Vest-Agder"
},
"RO": {
"name": "Rogaland"
},
"HO": {
"name": "Hordaland"
},
"SF": {
"name": "Sogn og Fjordane"
},
"MR": {
"name": "Møre og Romsdal"
},
"ST": {
"name": "Sør-Trøndelag"
},
"NO": {
"name": "Nord-Trøndelag"
},
"NT": {
"name": "Nordland"
},
"TR": {
"name": "Troms"
},
"FI": {
"name": "Finnmark"
},
"SJ": {
"name": "Svalbard"
},
"SJ": {
"name": "Jan Mayen"
},
"AK": {
"name": "Akershus"
},
"AA": {
"name": "Aust-Agder"
},
"BU": {
"name": "Buskerud"
},
"HE": {
"name": "Hedmark"
},
"OP": {
"name": "Oppland"
},
"OS": {
"name": "Oslo"
},
"TE": {
"name": "Telemark"
},
"VF": {
"name": "Vestfold"
},
"OF": {
"name": "Østfold"
}
}
What I am trying to do is to list the state names based on country selection in dropdown menu. Let say if Norway is selected as country, the dropdown will contain all the state names in list and short name in value
I created the code below, but getting empty dropdown
$conobjitems = json_decode($jsonitemfile);
echo "<select>";
$findBystatename = function($constatename) use ($conobjitems) {
foreach($conobjitems->Norway as $short){
echo "<option>". $short . "</option>";
}
};
echo "</select>";
Need advise here.

Answer 1

This is the correct code that I came up with
echo "<select>";
foreach($conobjitems->PK->states as $short => $outputstate) {
echo "<option value=". $short . ">". $outputstate->name . "</option>";
}
echo "</select>";