Member Avatar for xuexue

hi guys, i just want to ask, how could i get the value of the textbox
using the onchange() function. i know that using the form action and method get is one possible solution but what i want is that the value of the textbox in page1 should be passed to another php page page2 for some record hunting using database then pass it again to the page1 and put the result of the record hunting to another textbox..is that possible?

i know it is possible, but could you help me find the simplest possible way ever? tnx in advance..^^

  • 3 Contributors
  • 2 Replies
  • 15K Views
  • 4 Years Discussion Span
  • Latest Post Latest Post by Manoj_16

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Member Avatar for Shanti C

As i understood,

see this code:

<script type="text/javascript">
function change_send(x)
{
	//alert(x);
	window.href="page2.php?text="+x;
}
</script>
<html>
<p>
<table>
<tr>
<td>
Your full name:</td>
<td>
<input type="text" name="fullname" onChange="change_send(this.value);"></td>
</tr>
</table>
</html>

or come up with clear information.

Member Avatar for Manoj_16 
Using the below code i am getting the data from database(mysql) and displaying it in the textbox that is in the form of table(textbox inside the table data). Now I can edit the data(td) on browser(table) but the problem is updating it to database upon changing( onChange option ) data in that table and I want to display the updated data where i have edited. This i am trying to do in AJAX, PHP and MYSQL.Please do help me i am stuck here.Am i doing it in the right way?, I am a newbie. I dont want to use buttons to update.

edit.php

<html>
  <head>
    <script>


      function ajaxFunction(){
        var ajaxRequest;

        try{

          ajaxRequest = new XMLHttpRequest();
        }
        catch (e){
          alert("Something is wrong");
          return false;

        }

        ajaxRequest.onreadystatechange = function()
          {

            if(ajaxRequest.readyState == 4){

              var ajaxDisplay = document.getElementById('name');
              ajaxDisplay.innerHTML = ajaxRequest.responseText;


              var ajaxDisplay1 = document.getElementById('age');
              ajaxDisplay1.innerHTML = ajaxRequest.responseText;

            }
          }

          var id = document.getElementById('id').value;
        var name = document.getElementById('name').value;
        ;
        var age = document.getElementById('age').value;
        var queryString = "?id=" + id ;
        queryString +=  "&name=" + name + "&age=" + age;
        ajaxRequest.open("GET", "update.php" + queryString, true);
        ajaxRequest.send(null);


      }

      </script>

  </head>
  <body>

  <table>

      <tr>
        <th>
          ID
        </th>
      </th>
  <th>
    NAME
  </th>
  <th>
    AGE
  </th>
  <th>
  </th>
  </tr>

  <?php
include("database.php");

$query = "select * from person";

$sql = mysqli_query($conn,$query);

while($row = mysqli_fetch_array($sql))
{
?>

  <tr>

    <td>
      <input  name ="id" id="id"  type = "text" autocomplete="off" value=
      <?php echo $row['id'] ?>
      >
    </td>


    <td>
      <input  name="name" id="name"  type = "text" onChange = "ajaxFunction();" autocomplete="off" value=
      <?php echo $row['name'] ?>
      >
           </td>


           <td>
             <input  name ="age"  id="age" type = "text" onChange = "ajaxFunction();" autocomplete="off" value=
             <?php echo $row['age'] ?>
             >
           </td>
           <?php       
}       
?>

  </tr>
</table>
</body>

</html>




update.php

<?php
include('database.php');

$name = $age = $id ="";

if(isset($_GET['name']))
{$name=$_GET['name'];}

if(isset($_GET['age']))
{$age=$_GET['age'];}

if(isset($_GET['id']))
{$id=$_GET['id'];}


$query3 = "update person set name ='$name', age='$age' where id='$id'";

$sql1 = mysqli_query($conn,$query3);

if($sql1)

header('location:edit.php');

?>

Do help me ! Thansk in advance.
Edited by Manoj_16
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