How to explode json data already present in mysql database using php? - php

Question

I have a database table which previously had a coloumn data having a json data in it
| data | - `"{"display":{"pageName":"TEBT Home","Name":"rtryrtyrty rtyrtyrtyrty",
"Email":"trupti.gjklhjkgv#gmail.com","state":"WB","city":"CTY_ARAM",
"dob":"09/12/1964"}}"`
now I want to restructure my data_table and add different coloumns for pageName , Name, Email, state, city, dob and want to exlpode the json data already present in 'data' coloumn.
this is the logic I am trying to Implement , but it doesn't work .
<?php
$dataString = "SELECT data FROM data_table";
$result = mysqli_query($conn, $dataString);
$data =array();
while($row=mysqli_fetch_assoc($result)){
$data[] = $row['data'];
}
$iterations = ceil(count($data));
for ($x = 0; $x <$iterations; $x++) {
echo "The number is: ".$x." <br>";
if (!empty($data)) {
$json = json_decode($data, true);
$page = $json['display']['pageName'];
$name = $json['display']['Name'];
$email =$json['display']['Email'];
$mob =$json['display']['mobno'];
$state =$json['display']['state'];
$city =$json['display']['city'];
$dob =$json['display']['dob'];
echo $page;
$sql = "UPDATE data_table SET page = '".$page."' WHERE data ='".$data."' ";
mysqli_query($conn, $sql);
}
else
{
echo "done";
}
}
?>

Answer 1

use this code to see db error if any :
$dataString = "SELECT data FROM data_table";
$result = mysqli_query($conn, $dataString);
$data =array();
while($row=mysqli_fetch_assoc($result)){
$data[] = $row['data'];
}
$iterations = ceil(count($data));
for ($x = 0; $x <$iterations; $x++) {
echo "The number is: ".$x." <br>";
if (!empty($data)) {
$json = json_decode($data, true);
$page = $json['display']['pageName'];
$name = $json['display']['Name'];
$email =$json['display']['Email'];
$mob =$json['display']['mobno'];
$state =$json['display']['state'];
$city =$json['display']['city'];
$dob =$json['display']['dob'];
echo $page;
$sql = "UPDATE data_table SET page = '".$page."' WHERE data ='".$data."' ";
mysqli_query($conn, $sql);
echo("Error description: " . mysqli_error($con)); // added to see error msg
}