Fill the fields of a form from mysql query result

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.

What exactly you want dude? You want all value should be display or something else? — Kalpit
– Kalpit, Commented Sep 13, 2013 at 4:23
Thank you for the immediate response.. But, i didnt get ur reply exactly. — MNA Sha
– MNA Sha, Commented Sep 13, 2013 at 4:24
if the user enters designation code on designation.php and clicks update, you want to fill all form with the values from database corresponding to the designation code, right? — plain jane
– plain jane, Commented Sep 13, 2013 at 4:29
@Coder Thank you for reply. my need is, Fields like Designation, Salary Code, Salary Scale and etc in designation.php Should be filled automatically when the user enters the Designation Code. — MNA Sha
– MNA Sha, Commented Sep 13, 2013 at 4:29

Kalpit · Accepted Answer · 2013-09-13 04:53:22Z

2

  $(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...

edited Sep 13, 2013 at 4:53

answered Sep 13, 2013 at 4:48

Kalpit

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Ujjwal Ojha · Accepted Answer · 2013-09-13 04:59:07Z

1

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!!

answered Sep 13, 2013 at 4:59

Ujjwal Ojha

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