pass the id from one page to another

maideen · April 2, 2014

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
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

<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

<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';

edit.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>';
    
                }
    }

Tq

Maideen

cyberRobot · April 2, 2014

I don't see any $_POST variables, so perhaps the following will help:

http://www.tizag.com/phpT/postget.php

maideen · April 2, 2014

Hi Guru

This is the post action

Basically page shows all data as table format with id. If i click the selected update button, the id of selected row to be posted into next pages to retrive the data of the particular Id. Can you pls explain how to do?

<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
}

cyberRobot · April 2, 2014

It looks like you have one form which contains many input fields named "vid". Unless I'm missing something, that's not going to work. When the form is submitted, only the last value will be passed through the form.

Instead, I would recommend looking into using anchor tags for the action buttons. For example, you could try something like the following:

<?php
if(isset($result)){
     while($row = mysqli_fetch_array($result)){
          echo '<tr>';
          echo "<td>{$row['id']}</td>";
          echo "<td>{$row['name']}</td>";
          echo "<td>{$row['firstname']}</td>";
          echo "<td>{$row['age']}</td>";
          echo "<td><a href='data.php?update=1&vid={$row['id']}'>Update</a> <a href='data.php?delete=1&vid={$row['id']}'>Delete</a></td>";
          echo '</tr>';
     }
}
?>

Then on the data.php page, you could see which link was clicked by doing something like this:

<?php
if(isset($_GET['update'])) {
     //do update stuff
} elseif(isset($_GET['delete'])) {
     //do delete stuff
}
?>

Edited April 2, 2014 by cyberRobot

maideen · April 3, 2014

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

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

Ch0cu3r · April 3, 2014

if you are setting up the UPDATE link as cyberRobot suggested then you can retrieve the users id using $_GET['vid']

if(isset($_GET['update']))
{
   $id = intval($_GET['vid']); // get the user id, sanitize it before using it an SQL query
   $sql="select name, firstname, age from customer where id=$id"; // select name, firstname and age fields, no need to get id field again as we already have it ($id)
   ...
}

maideen · April 3, 2014

hi Ch0cu3r

When I use below code, i got id of last record. If I have 100 record in database, I shows last record i.e 100. But I need what I selected. Pls help me

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

<?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';  
            //header("Location:edit.php"); 
     }

Edited April 3, 2014 by maideen

Ch0cu3r · April 3, 2014

Re-read cyberRobots reply (import bits are in bold)

It looks like you have one form which contains many input fields named "vid". Unless I'm missing something, that's not going to work. When the form is submitted, only the last value will be passed through the form.

Instead, I would recommend looking into using anchor tags for the action buttons.

Either do as cyberRobot suggested. Or change your code so it outputs a new form for each row.

.

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pass the id from one page to another

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