select all rows and output into arrays phpmyadmin

Question

hey there im a little stuck with getting data from my phpmyadmin and displaying it as json this is my php so far

<?php

$user = '';
$password = '';
$con=mysqli_connect("localhost",$user,$password,"tugutept_jumputi");
if (mysqli_connect_errno())
  {
   echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }
$query = 'SELECT * FROM chartable';


$stm = mysqli_query($con,$query);

$data_array = array();
$count = 0;
while($r = mysqli_fetch_row($stm)){
    $data_array[$count] = $r;
    $count= $count +1;

}
echo json_encode($data_array);
mysqli_close($con);


?>

which displays this when i want it looking like this

[
  {
    "ID": "000001"
    "Name": "[The Saiyan who grew up on Earth] Son Goku",
    "Portrait": "BaseGoku.png",
    "Series": "Dragon Ball",
    "MaxRarity": "6 start",
    "Type": "Blue",
    "Class": "Tank",
    "Era": "1980",
    "Release Date": "2018-03-28",
    "Farmable": "0",
    "Method": "Trade Medal Store"
  }
]

i want to take each row from the table and put it in an array but i keep getting the whole table in one array, as you can see im a little stumped any help is welcome thanks

thanks i have been trying to put a foreach loop but i keep getting a 500 error not to sure if this is correct — Mrtcupz, Jan 28 '20 at 17:38
$query = 'SELECT * FROM chartable'; $stm = mysqli_query($con,$query); $data_array = array(); $count = 0; foreach($data_array as $v){ $data_array[$count] => $v; $count= $count +1; } echo json_encode($data_array); mysqli_close($con); — Mrtcupz, Jan 28 '20 at 17:38
Please read https://stackoverflow.com/questions/58808332/should-we-ever-check-for-mysqli-connect-errors-manually — Dharman, Jan 28 '20 at 21:22
Could you please update the question to show us what changed you have made. Could you also explain what exactly the issue is? I don't understand it from the description. — Dharman, Jan 28 '20 at 21:23
Just changed the while loop to a foreach loop but i cant get it to work — Mrtcupz, Jan 28 '20 at 22:35

score 0 · Answer 1 · answered Jan 28 '20 at 07:25

At the moment, you are fetching the data with a numeric index using mysqli_fetch_row() - which says from the manual

Get a result row as an enumerated array

As the json_encode() sees a sequential numeric array starting at 0, it will encode it as an array and not an object.

If you want the column names, change that to mysqli_fetch_assoc()...

while($r = mysqli_fetch_assoc($stm)){

Which says

Fetch a result row as an associative array

select all rows and output into arrays phpmyadmin

1 Answers1