Updating Information in a MySQL Database Using PHP. Please help! Haha :)

[Transmission94]

Hi there, I have appropriated a script from http://www.spoono.com/php/tutorials/tutorial.php?id=23 for my own needs, but it is not working out for me. I have 3 PHP pages all in the same directory (Please not that the comments were not all written by me and thus might not be properly describing what is happening):

 

1.php:

<?php 
//connect to mysql
//change user and password to your mySQL name and password
mysql_connect("localhost","admin","password"); 

//select which database you want to edit
mysql_select_db("shop"); 

//If cmd has not been initialized
if(!isset($cmd)) 
{
   //display all the shit
   $result = mysql_query("select * from product order by prod_id") or die(mysql_error()); 
   
   //run the while loop that grabs all the products
   while($r=mysql_fetch_array($result)) 
   { 
      //grab the title and the ID of the news
      $prod_name=$r["prod_name"];//take out the product name
      $prod_id=$r["prod_id"];//take out the id
     
 //make the title a link
      echo "<a href='2.php?cmd=edit&prod_id=$prod_id'>$prod_name - Edit</a>";
      echo "<br>";
    }
}
?>

 

 

 

2.php:

<?php

mysql_connect("localhost","admin","password") or die ("no connection"); 

//select which database you want to edit
mysql_select_db("shop") or die ("no shop");


if($_GET["cmd"]=="edit" || $_POST["cmd"]=="edit")
{
   if (!isset($_POST["submit"]))
   {
      $prod_id = $_GET["prod_id"];
      $sql = "SELECT * FROM product WHERE prod_id=$prod_id";
      $result = mysql_query($sql) or die ("Oops");        
      $myrow = mysql_fetch_array($result); 
   }
      ?>
  
      <form action="3.php" method="post">
      <input type=hidden name="prod_id" value="<?php echo $myrow["prod_id"] ?>">
   
      Product Name:<INPUT TYPE="TEXT" NAME="prod_name" VALUE="<?php echo $myrow["prod_name"] ?>" SIZE=30><br>
      Description:<TEXTAREA NAME="prod_description" ROWS=10 COLS=30><?php echo $myrow["prod_description"] ?></TEXTAREA><br>
      Price:<INPUT TYPE="TEXT" NAME="prod_price" VALUE="<?php echo $myrow["prod_price"] ?>" SIZE=30><br>
   
      <input type="hidden" name="cmd" value="edit">
   
      <input type="submit" name="submit" value="submit">
   
      </form>
   
   <?php } ?>

 

 

3.php

<?php
   if ($_POST["$submit"])
   {
      $prod_name = $_POST["prod_name"];
      $prod_price = $_POST["prod_price"];
  $prod_description = $_POST["prod_description"];

      $sql = "UPDATE product SET prod_name='$prod_name',prod_price='$prod_price',prod_description='$prod_description' WHERE prod_id=$prod_id";

      $result = mysql_query($sql);
      echo "Thank you! Information updated.";
   }
else die("<h1>FOR THE LOVE OF GOD, WORK!</h1>");
?>

 

 

Basically, the first two pages do what they are supposed to, but in 3.php the condition of ($_POST["$submit"]) is not met, and thus the code underneath it isn't executed. I don't have a good enough understanding to know why the condition is not being met.

 

Any help is very much appreciated!

 

 

Thanks,

 

 

 

Transmission94

[mrMarcus]

looks like a typo .. change:

 

if ($_POST["$submit"])

 

to:

 

if ($_POST["submit"])

 

notice i removed the $

[Transmission94]

Hi mrMarcus!

 

Thanks for the input, but when I do as you say I get the error...

Warning: mysql_query() [function.mysql-query]: Access denied for user 'ODBC'@'localhost' (using password: NO) in D:\xampp\htdocs\3.php on line 19

Warning: mysql_query() [function.mysql-query]: A link to the server could not be established in D:\xampp\htdocs\3.php on line 19
Thank you! Information updated. 

... instead of my own error message.

[mrMarcus]

now yo have to include your database connection to 3.php and you will be set.

 

your best bet is to take this:

 

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

 

and put it in a file called connect.php (or whatever you like), and include it within it page thereafter that requires a database connection.

 

so, for 2.php, you would replace:

 

mysql_connect("localhost","admin","password") or die ("no connection"); 
   
//select which database you want to edit
mysql_select_db("shop") or die ("no shop");

 

with:

 

include ('/path/to/connect.php'); //path/to/ of course being the path to your file .. if it resides within the same folder for test purposes, just have connect.php with no path;