Compare variable of two different array? - php

Question

Here i want compare two array.
I have two tables one is CATEGORY and second is USER(in which selected multiple category is stored in this format , , , ).
but when the USER want to update their category at that time the earlier selected category should return checked 1 and which is not selected should return checked 0 and here is my code.
case 'cat':
$sql="SELECT category from nesbaty_user where user_id='".$user_id."'";
$qry_res=mysqli_query($con,$sql);
$response1 = array();
while ($array = mysqli_fetch_array($qry_res))
{
foreach (explode(',', $array['category']) as $cat)
{
$response1[]=array('category' => $cat);
$response['Selected_category'] = $response1;
}
}
$qry="SELECT cat_id,category,image_url FROM nesbaty_category";
$qry_res=mysqli_query($con,$qry);
$jsonData = array();
while ($array = mysqli_fetch_assoc($qry_res))
{
$jsonData[] = $array;
$response['category'] = $jsonData;
}
echo json_encode($response);
//echo json_encode(array('data1' =>$response1));
//
mysqli_close($con);
break;
and here is my fetched array from the database.
{
"Selected_category": [
{
"category": "5"
},
{
"category": "6"
},
{
"category": "9"
}
],
"category": [
{
"cat_id": "1",
"category": "Drug",
"image_url": "1.jpg"
},
{
"cat_id": "2",
"category": "Bars",
"image_url": "2.jpg"
},
{
"cat_id": "3",
"category": "Bars",
"image_url": "2.jpg"
},
{
"cat_id": "4",
"category": "Hair Saloon",
"image_url": "2.jpg"
}
]
}

Answer 1

This is for handle this same structure as You provided.
<?php
$response = [];
$sql = "SELECT category from nesbaty_user where user_id='".$user_id."'";
$qry_res = mysqli_query($con, $sql);
$selectedCategories = mysqli_fetch_array($qry_res);
$selectedCategories = explode(',', $selectedCategories['category']);
$qry = "SELECT cat_id,category,image_url FROM nesbaty_category";
$qry_res = mysqli_query($con, $qry);
while ($categories = mysqli_fetch_assoc($qry_res)) {
$categories['checked'] = (int)\in_array($categories['cat_id'], $selectedCategories);
$response[] = $categories;
}
echo json_encode($response);
mysqli_close($con);
If any error writes a comment, I'll fix it.