PHP SQL Edit Form

johneva posted this at 11:26 — 12th February 2010.

Joined: Jan 2008

could someone take a look over this for me and see where I have gone wrong I carnt spot what I have done wrong I am new at this an its driving me mad now.

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">

<head>
<link href="cp_css/loginmodule.css" rel="stylesheet" type="text/css" />
</head>
  
  <body>
   <h1>Edit Form</h1>
   <p><a href="http://localhost/dual_control_cars/control_panel">Control Panel</a></p>

<ul>
<?php 
//connect to mysql
//change user and password to your mySQL name and password
mysql_connect("localhost", "johnmev_user1", "Pass1");
    
//select which database you want to edit
mysql_select_db("cp"); 

//If cmd has not been initialized
if(!isset($cmd)) 
{
   //display all the cars
   $result = mysql_query("select * from data order by id"); 
   
   //run the while loop that grabs all the cars
   while($r=mysql_fetch_array($result)) 
   { 
      //grab the model and the ID
      $make=$r["make"];//take out the model     
      $model=$r["model"];//take out the model
      $id=$r["id"];//take out the id
     
     //show the make and model a link in a list
      echo "<li>";
      echo "<a href='edit.php?cmd=edit&id=$id'>Edit - $make $model</a>";
      echo "</li>";
    }
}
?>
   </ul>

<?php
if($_GET["cmd"]=="edit" || $_POST["cmd"]=="edit")
{
   if (!isset($_POST["submit"]))
   {
      $id = $_GET["id"];
      $sql = "SELECT * FROM data WHERE id=$id";
      $result = mysql_query($sql);        
      $myrow = mysql_fetch_array($result);
?>
 
      <form action="edit.php" method="post">
      <input type=hidden name="id" value="<?php echo $myrow["id"] ?>">
   
      Make:<input type="text" name="make" value="<?php echo $myrow["make"] ?>" size=30 /><br />
      Model:<input type="text" name="model" value="<?php echo $myrow["model"] ?>" size=30 /><br />
      Spec:<input type="text" NAME="spec" value="<?php echo $myrow["spec"] ?>" size=30 /><br />
      7 Month Rental:<input type="text" NAME="seven" value="<?php echo $myrow["seven"] ?>" size=30 /><br />
      12 Month Rental:<input type="text" NAME="twelve" value="<?php echo $myrow["twelve"] ?>" size=30 /><br />
      18 Month Rental:<input type="text" NAME="eighteen" value="<?php echo $myrow["eighteen"] ?>" size=30 /><br />
   
      <input type="hidden" name="cmd" value="edit" />
   
      <input type="submit" name="submit" value="submit" />
   
      </form>

     <?php } 
     
   if ($_POST["$submit"])
   {
      $make = $_POST["make"];
      $model = $_POST["model"];
      $spec = $_POST["spec"];
      $seven = $_POST["seven"];
      $twelve = $_POST["twelve"];
      $eighteen = $_POST["eighteen"];


      $sql = "UPDATE data SET make='$make',model='$model',spec='$spec',seven='$seven',twelve='$twelve',eighteen='$eighteen' WHERE id=$id";

      $result = mysql_query($sql);
      echo "Thank you! Information updated.";
   }
}
?>

</body>
</html>

pr0gr4mm3r posted this at 14:49 — 13th February 2010.

He has: 1,502 posts

Joined: Sep 2006

So...what's the problem?

cmoyer posted this at 02:41 — 21st February 2010.

He has: 131 posts

Joined: Jun 2008

I think I see a minor problem in your inline php
ex:

Make:<input type="text" name="make" value="<?php echo $myrow["make"] ?>" size=30 /><br />

you need a semi colon after the echo $myrow["make"]
ex:

Make:<input type="text" name="make" value="<?php echo $myrow["make"]; ?>" size=30 /><br />

Hope that helps

sandeep Kumar posted this at 06:42 — 13th July 2010.

He has: 53 posts

Joined: Jun 2010

Hi Johneva,
I corrcet your code, You Just try with them,

Edit Form
Control Panel

<?php
if(isset($_POST['cmd']) == "edit")
{
//connect to mysql
//change user and password to your mySQL name and password
$connection=mysql_connect("localhost", "johnmev_user1", "Pass1");
   
//select which database you want to edit
mysql_select_db("cp",$connection);

//If cmd has not been initialized

   //display all the cars
   $result = mysql_query("select * from data order by id",$connection);
  
   //run the while loop that grabs all the cars
   while($r=mysql_fetch_array($result))
   {
      //grab the model and the ID
      $make=$r["make"];//take out the model    
      $model=$r["model"];//take out the model
      $id=$r["id"];//take out the id
    
     //show the make and model a link in a list
      echo "<li>";
      echo "<a href='edit.php?cmd=edit&id=$id'>Edit - $make $model</a>";
      echo "</li>";
    }
}
?>

<?php
if(isset($_POST['cmd']) == "edit")
{
   
      $id = $_GET["id"];
      $sql = " SELECT * FROM data WHERE id = '$id' ";
      $result = mysql_query($sql);       
      $myrow = mysql_fetch_array($result);
?>

Make:
Model:
Spec:
7 Month Rental:
12 Month Rental:
18 Month Rental:

<?php
 }
    
   if ($_POST["$submit"])
   {
      $make = $_POST["make"];
      $model = $_POST["model"];
      $spec = $_POST["spec"];
      $seven = $_POST["seven"];
      $twelve = $_POST["twelve"];
      $eighteen = $_POST["eighteen"];


      $sql = ("UPDATE data SET                         
                              make='$make',
                              model='$model',
                              spec='$spec',
                              seven='$seven',
                              twelve='$twelve',
                              eighteen='$eighteen' 
                              WHERE id='$id' ", $connection );

      $result = mysql_query($sql);
      echo "Thank you! Information updated.";
   }
}
?>

and still you face problem then, you mention here. & I'll try to solve them, now Enjoy with that code. ):

Greg K posted this at 15:29 — 13th July 2010.

He has: 2,145 posts

Joined: Nov 2003

One important note that is before this were to go into production (intended for public use), you need to validate/cleanse the data coming into $_POST.

(think what would happen if I entered the following for MAKE:

hello'; drop table data;

Now when it submits, the code as is will delete your entire table.

-Greg