PHP script only display one row of data in android studio - php

Question

I would like to ask how can I display many rows of data instead of only one row. The following code will only display one row of record instead of multiple records are available. I'm using mysqli_prepare statements here. Or the problem is on my android studio coding? My application is implemented with login function and coding as below.
<?php
$host="DB_HOST";
$user="DB_USER";
$password="DB_PASSWORD";
$db="DB_NAME";
$con = mysqli_connect($host,$user,$password,$db);
$parentic=$_POST["ParentIC"];
$password=$_POST["Password"];
$selectquery = mysqli_prepare($con, "SELECT Parents_Data.Name, Student_List.StudName, Student_List.StudIC, Student_List.Form,Student_List.Class,discipline_record.Date,discipline_record.RulesCode, discipline_record.TypesofMistakes,discipline_record.Punishment FROM discipline_record LEFT JOIN Student_List ON discipline_record.StudIC = Student_List.StudIC LEFT JOIN Parents_Data ON Student_List.ParentIC = Parents_Data.ParentIC WHERE Parents_Data.ParentIC = ? AND Parents_Data.Password = ? ");
mysqli_stmt_bind_param ($selectquery, "ss", $parentic, $password);
mysqli_stmt_execute($selectquery);
mysqli_stmt_store_result($selectquery);
mysqli_stmt_bind_result($selectquery,$name,$studname,$studic,$form,$classs,$ddate,$code,$mistakes,$punishment);
$user = array();
while(mysqli_stmt_fetch($selectquery))
{
$user[name]=$name;
$user[studname]=$studname;
$user[studic] = $studic;
$user[form]=$form;
$user[classs]=$classs;
$user[ddate]=$ddate;
$user[code]=$code;
$user[mistakes]=$mistakes;
$user[punishment]=$punishment;
}
echo json_encode($user);
mysqli_stmt_close($selectquery);
mysqli_close($con);
?>

Answer 1

I would go with something like that:
$userGroup = array();
$user = array();
while(mysqli_stmt_fetch($selectquery))
{
$user[name]=$name;
$user[studname]=$studname;
$user[studic] = $studic;
$user[form]=$form;
$user[classs]=$classs;
$user[ddate]=$ddate;
$user[code]=$code;
$user[mistakes]=$mistakes;
$user[punishment]=$punishment;
array_push($userGroup,$user);
}
echo json_encode($userGroup);

Answer 2

It might be easier to use 2-dimensional array
$users = array();
while(mysqli_stmt_fetch($selectquery))
{
$users[] = array();
$users[][name]=$name;
$users[][studname]=$studname;
$users[][studic] = $studic;
$users[][form]=$form;
$users[][classs]=$classs;
$users[][ddate]=$ddate;
$users[][code]=$code;
$users[][mistakes]=$mistakes;
$users[][punishment]=$punishment;
}
foreach ( $users as $user )
echo json_encode($user);

Answer 3

It looks like you are doing correct query, but when fetching results using one flat array instead two dimension.
Also, try to avoid publishing your DB credentials in public :) It is quite dangerous.
So you should do something like that:
<?php
$host="YOUR_DB_HOST";
$user="YOUR_DB_USER";
$password="YOUR_DB_USER_PASSWORD";
$db="YOUR_DB_NAME";
$con = mysqli_connect($host,$user,$password,$db);
$parentic=$_POST["ParentIC"];
$password=$_POST["Password"];
$selectquery = mysqli_prepare($con, "SELECT Parents_Data.Name, Student_List.StudName, Student_List.StudIC, Student_List.Form,Student_List.Class,discipline_record.Date,discipline_record.RulesCode, discipline_record.TypesofMistakes,discipline_record.Punishment FROM discipline_record LEFT JOIN Student_List ON discipline_record.StudIC = Student_List.StudIC LEFT JOIN Parents_Data ON Student_List.ParentIC = Parents_Data.ParentIC WHERE Parents_Data.ParentIC = ? AND Parents_Data.Password = ? ");
mysqli_stmt_bind_param ($selectquery, "ss", $parentic, $password);
mysqli_stmt_execute($selectquery);
mysqli_stmt_store_result($selectquery);
mysqli_stmt_bind_result($selectquery,$name,$studname,$studic,$form,$classs,$ddate,$code,$mistakes,$punishment);
$user = array();
$users = array();
while(mysqli_stmt_fetch($selectquery))
{
$user[name]=$name;
$user[studname]=$studname;
$user[studic] = $studic;
$user[form]=$form;
$user[classs]=$classs;
$user[ddate]=$ddate;
$user[code]=$code;
$user[mistakes]=$mistakes;
$user[punishment]=$punishment;
$users[] = $user;
}
mysqli_stmt_close($selectquery);
mysqli_close($con);
echo json_encode($users);