how to post id to another page

Question

5.00/5 (1 vote)

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Hi I am new, Pls help me the following issue. When click the update button, ID should be to another page to process the update function. Below is my entire code

index.php

PHP

<?php
include_once '../../inc/config.inc.php';

$sql="SELECT * FROM demo  order by id";
    try {
            $result = mysqli_query($con,$sql);  
            include 'view.html.php'; 
         } 
    catch (PDOException $e) 
        {
           echo $e->getMessage() . "\n";
           file_put_contents('PDOErrors.txt', $e->getMessage(), FILE_APPEND);
//           $output = 'Error fetching authors from database!';
//           include '../../notification/errormsg.php';
           exit();
        }

view.html.php

PHP

<Script Language="javascript">
function change_action()
    {
        var frm_obj=document.getElementById("frm");
        frm_obj.action="data.php";
    }
</Script>

<h3>User Details</h3>
<form action="" method="POST" id="frm" >
    <table width="100%" align="center" cellpadding="4" cellspacing="1">
        <tr>
            
            <td>ID</td>
            <td>ID</td>
            <td>NAME</td>
            <td>FIRST NAME</td>
            <td>AGE</td>
            <td></td>
            
        </tr>
        <?php
            if(isset($result)){
            while($row = mysqli_fetch_array($result)){ ?>
             <tr>
                
                <td><input type="text" name="vid" value="<?php echo $row['id'];?>"/></td>
                <td><?php echo $row['id'];?></td>
                <td><?php echo $row['name'];?></td>
                <td><?php echo $row['firstname'];?></td>
                <td><?php echo $row['age'] ;?></td>
                <td><input type="submit" value="update" name="update" onclick="change_action()">
                <input type="submit" value="delete" name="delete" onclick="change_action()">
                </td>                
              </tr>  
         <?php
            } 
         
            }
            mysqli_close($con);
        ?>
   
    </table>
</FORM>
<?php
include '../../inc/footer.php';

data.php

PHP

<?php
include_once '../../inc/config.inc.php';
if (isset($_POST['update']) && $_POST['update']  != "" )
        {
            $id = $_POST['vid'];
            $sql="SELECT * FROM demo where id='$id'";
            $result = mysqli_query($con,$sql);
            include 'edit.php';
     }

edit.php

PHP

<?php
echo 'edit.php';
  
if(isset($result))
    {
        while($row = mysqli_fetch_array($result))
                { 
                    echo $row['id'];'<br>';
                    echo $row['name'];'<br>';
                    echo $row['firstname'];'<br>';
                    echo $row['age'];'<br>';
    
                }
    }

thank you

maideen

Posted 2-Apr-14 2:12am

Maideen Abdul Kader

Updated 30-May-22 7:37am

Add a Solution

Comments

Mohibur Rashid 3-Apr-14 1:44am

look up action

5 solutions

Solution 4

You can solve your problem by using Jquery

Give id to each element which holds id like

HTML

<table>
   <tr id="r1">
      <td id="idn1">1</td>
      <td>Hoffman</td>
      <td>Tatyana</td>
      <td>100</td>
      <td onclick="update('idn1',1)">Update</td>
   </tr>

   <tr id="r1">
      <td id="idn2">2</td>
      <td>Atkins</td>
      <td>Ishmael</td>
      <td>44</td>
      <td onclick="update('idn2',2)">Update</td>
   </tr>
   ...
   ...
   ...
   ...
   ...
</table>

Now write the javascript function

JavaScript

function update(idn,n)
{
  cid=$('#'+idn).val();
  $.post("your_next.php?idn="+cid,function(data){
    $("r"+n).html(data);
  });
}

In "your_next.php" you have the id to whom you want to update values.....

Posted 6-Apr-14 23:18pm

kishore@cse

Solution 2

I am telling the scnario. I have page with all customer data as grid format. last column is "UPDATE". If I select 5th row and click update, next page open the details of 5th row customer data.
My sql command:

$sql="select id,name,firstname,age from customer where id='$id'"

How can i pass the id for select customer to the next page to retrive data from mysql database and to update if any changes.

Pls help me

HTML

ID  NAME          FIRST NAME    AGE
1   Hoffman       Tatyana       100     Update
2   Atkins        Ishmael        44     Update
3   Hamilton      Mohammad       73     Update
4   Murray        Troy       18     Update
5   Schwartz      Carla      79     Update
6   Bond          Leno       33     Update
7   Noble         Georgia        88     Update
8   Frederick     Bradley        80     Update
9   Chambers      Aphrodite      63     Update
10  Schultz       Allegra        90     Update
11  Garrett       Lillith        20     Update
12  Mcdaniel      Venus      89     Update
13  Acevedo       Iola       26     Update

Posted 3-Apr-14 1:29am

Maideen Abdul Kader

Solution 5

visit here..............

How to pass image and values to another html page at one button click in html[^]

Posted 6-Apr-14 23:34pm

Rakesh Meel

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Mr_Programmer31 · Accepted Answer · 2014-04-02T22:22:00

Solution 1

<a href="data.php?id=1">Update</a>


on data.php

$myId = $_GET['id'];

Posted 2-Apr-14 22:22pm

Mr_Programmer31

Updated 2-Apr-14 22:23pm

v2

Member 10724212 · Accepted Answer · 2014-04-04T00:17:00

Solution 3

you can just put link lines as below:

Update

on data.php

$myId = $_GET['id'];

or use

@$_GET['id'];

Posted 4-Apr-14 0:17am

Member 10724212