Fill the fields of a form from mysql query result - php

Question

In my project it has an interface(designation.php) which has a form. When a user enters a 'designation code'(which is a primary key) and click the 'update' button, the
remaining fields should be filled accordingly. I understand that Ajax will be needed for this.
but the problem is, i'm done with retrieving the result only as a single row from the database by using following query. but, this is not my objective.
while($row = mysqli_fetch_assoc($run1)) {
echo "<tr id='tr'>";
foreach($row AS $cell) echo "<td>$cell</td>";
echo "</tr>\n";
}
Following is the 'designation.php'..
<form action="crud.php" method="post">
<table>
<th >DESIGNATION</th>
<tr>
<td >Designation Code</td>
<td ><input type="text" name="des_code" /></td>
</tr>
<tr>
<td >Designation</td>
<td><input type="text" name="desig" /></td>
</tr>
<tr>
<td >Salary Code</td>
<td><input type="text" name="s_code" /></td>
</tr>
<tr>
<td >Salary Scale</td>
<td><input type="text" name="s_scale" /></td>
</tr>
<tr>
<td >Salary point </td>
<td><input type="text" name="s_point" /></td>
</tr>
<tr>
<td>Date</td>
<td > <input type="date" name="date" /></td>
</tr>
<tr>
<td><input type="submit" name="update" value="Update" /></td>
</tr>
</table>
I attempted many solutions but have not gotten the intended result. Any help is appreciated.

Answer 1

$(document).ready(function(){
("#form1").submit(function({
var id=<?php echo $_POST['des_code']; ?>;
$.ajax({
url: 'update.php',
type:'POST',
data: {id:id},
success : function (data)
{
var a=data.split('[BRK]');
$("#s_scale").val(a[0]);
// like wise store all your fields..
}
})
})
upadte.php will contain
$id=$_POST['id']; //designation id from jqyery
$q=mysqli_query("your goes here..");
while($row = mysqli_fetch_array($q))
{
$cell=$row[0];
$cell.="[BRK]".$row[1];
$cell.="[BRK]".$row[2];
}
echo $cell;
give id to your text field so from jquery you can add value directly.. and give form id also so you can fire onsubmit event...
i didn't try this code but hope it will work fine...

Answer 2

To transfer more than one value use json!
Jquery Code::
$(function(){
id="XXXX";
$("form").on("submit",function(e){
e.preventDefault();
$.getJson("update.php",{"id":id},function(response){
$("#value1").val(response.value1);
$("#value2").val(response.value2);
// like wise store all your fields..
})
})
})
Update.php::
$q=mysqli_query("your query");
$row = mysqli_fetch_array($q);
echo json_encode($row);
$row will be an array containing value1, value2 and so on!!