Updating a mysql record error

ipodman · April 12, 2009

Hey guys,

Well I have been trying to sort this for the past hour and I have been searching for the solution but if I missed it, it because I am tired.

I am trying to update a record in mysql but when I click submit and it goes to update.php it states it has been updated when it has now.

here is the form that calls that items from the database

<?





if (isset($_GET['pro_id']))
  $pro_id = $_GET['pro_id'];




$db= mysql_connect("localhost", "root");
mysql_select_db("site",$db);

$sql = "SELECT pro_id, name, sale_price, cat_id, description, stock_level FROM stock WHERE pro_id = $pro_id";
$result = mysql_query($sql) or die(mysql_error());

$onlyRecord = mysql_fetch_array($result);
$name = $onlyRecord['name'];
$sale = $onlyRecord['sale_price'];
$cat = $onlyRecord['cat_id'];
$id = $onlyRecord['description'];
$level = $onlyRecord['stock_level'];
  
?>

<html>
<head><title>Edit Record</title></head>
<body>
<form action="update.php" method="POST">
<input type="hidden" value="<? echo $pro_id; ?>">

<table>
<tr><td>ID</td><td><input type="text" name="sitename" value="<? echo $pro_id; ?>"></td></tr>

<tr><td>Name</td><td><input type="text" name="sitename" value="<? echo $name; ?>"></td></tr>

<tr><td>Sale Price</td><td><input type="text" name="sitename" value="<? echo $sale; ?>"></td></tr>
<tr><td>Category ID</td><td><input type="text" name="sitename" value="<? echo $cat; ?>"></td></tr>
<tr><td>description</td><td><input type="text" name="sitename" value="<? echo $id; ?>"></td></tr>
<tr><td>Stock Level</td><td><input type="text" name="sitename" value="<? echo $level; ?>"></td></tr>

</table>

<br />
<input type="submit" value="Update">
</form>
<a href="viewedit.php" target="_self">Go back without saving any changes</a>
</body>
</html>

And now the update.php which updates the database

<?php
$db= mysql_connect("localhost", "root", "");
mysql_select_db("site");

if (isset($_GET['pro_id']))
$pro_id = $_GET['pro_id'];
$name = $_GET['name'];
$sale = $_GET['sale'];
$cat = $_GET['cat'];
$id = $_GET['id'];
$level = $_GET['level'];
  


$sql= ("UPDATE stock SET name='$name', sale_price='$sale', description='$id', stock_level='$level' WHERE pro_id='$pro_id'");
mysql_query($sql) or die(mysql_error());

echo "Record submitted successfully.";
mysql_close();
?>
  
<meta http-equiv="Refresh" content="1; url=viewedit.php"><title>Remove Item</title>

Any help would be good, thanks

jackpf · April 12, 2009

What error are you getting? And what, does it just not update it? Does it wipe the record? Little more info would be cool...

ipodman · April 12, 2009

Hey,

It states that the update has happened but does not update, no changes are made to the record

jackpf · April 12, 2009

mysql_select_db("site",$db);

What's going on with that?

I'm pretty sure mysql_select_db() only accepts one database.

ipodman · April 12, 2009

mysql_select_db("site",$db);

What's going on with that?

I'm pretty sure mysql_select_db() only accepts one database.

I have just edited my first post to show new changes to the code [2nd code] but still have the same error of it saying it has been updated when its not.

jamesxg1 · April 12, 2009

<?php

$db = mysql_connect("localhost", "root", "");
mysql_select_db("site,$db");

if (isset($_GET['pro_id']))
$pro_id = $_GET['pro_id'];
$name = $_GET['name'];
$sale = $_GET['sale'];
$cat = $_GET['cat'];
$id = $_GET['id'];
$level = $_GET['level'];
  


$sql= ("UPDATE `stock` SET name = '$name', sale_price = '$sale', description = '$id', stock_level = '$level' WHERE pro_id = '$pro_id'") or die(mysql_error());
mysql_query($sql) or die(mysql_error());

echo "Record submitted successfully.";
mysql_close();
?>

<meta http-equiv="Refresh" content="1; url=viewedit.php"><title>Remove Item</title>

Try that ^^ and tell me what you get ?

jackpf · April 12, 2009

You should check for a mysql_error() after everything though - connections, selecting databases.

I don't know how many times I've said this over the last couple of days...

ipodman · April 12, 2009

<?php

$db = mysql_connect("localhost", "root", "");
mysql_select_db("site,$db");

if (isset($_GET['pro_id']))
$pro_id = $_GET['pro_id'];
$name = $_GET['name'];
$sale = $_GET['sale'];
$cat = $_GET['cat'];
$id = $_GET['id'];
$level = $_GET['level'];
  


$sql= ("UPDATE `stock` SET name = '$name', sale_price = '$sale', description = '$id', stock_level = '$level' WHERE pro_id = '$pro_id'") or die(mysql_error());
mysql_query($sql) or die(mysql_error());

echo "Record submitted successfully.";
mysql_close();
?>

<meta http-equiv="Refresh" content="1; url=viewedit.php"><title>Remove Item</title>

Try that ^^ and tell me what you get ?

Using the code you said i get the error of no database so i change the database code to:-

$db = mysql_connect("localhost", "root", "");
mysql_select_db("site");

then i go to change a record an dget told it has been changed when it still has not

jamesxg1 · April 12, 2009

ok its the top code you posted, let me have a butchers and ill see what i can do.

jamesxg1 · April 12, 2009

just a question, on your Edit Record form does it actually put the mysql records in the value of the text fields ?

jackpf · April 12, 2009

Jesus christ...

ipodman · April 12, 2009

just a question, on your Edit Record form does it actually put the mysql records in the value of the text fields ?

Yes, it shows the correct data taken from the database

Also jackpf I added th emysql error code and nothing came up

jamesxg1 · April 12, 2009

just a question, on your Edit Record form does it actually put the mysql records in the value of the text fields ?

Yes, it shows the correct data taken from the database

Also jackpf I added th emysql error code and nothing came up

ok no problem

try this form for your Edit Record page

<?php

if (isset($_GET['pro_id']))
  $pro_id = mysql_real_escape_string($_GET['pro_id']);


$db= mysql_connect("localhost", "root");
mysql_select_db("site");

$sql = "SELECT `pro_id`, `name`, `sale_price`, `cat_id`, `description`, `stock_level` FROM stock WHERE pro_id = '$pro_id'" or die(mysql_error());
$result = mysql_query($sql) or die(mysql_error());

$onlyRecord = mysql_fetch_array($result);

$name = $onlyRecord['name'];
$sale = $onlyRecord['sale_price'];
$cat = $onlyRecord['cat_id'];
$id = $onlyRecord['description'];
$level = $onlyRecord['stock_level'];

?>

<html>
<head><title>Edit Record</title></head>
<body>
<form action="update.php" method="POST">
<input type="hidden" value="<?php print $pro_id; ?>">

<table>
<tr><td>ID</td><td><input type="text" name="sitename" value="<?php print $pro_id; ?>"></td></tr>

<tr><td>Name</td><td><input type="text" name="sitename" value="<?php print $name; ?>"></td></tr>

<tr><td>Sale Price</td><td><input type="text" name="sitename" value="<?php print $sale; ?>"></td></tr>
<tr><td>Category ID</td><td><input type="text" name="sitename" value="<?php print $cat; ?>"></td></tr>
<tr><td>description</td><td><input type="text" name="sitename" value="<?php print $id; ?>"></td></tr>
<tr><td>Stock Level</td><td><input type="text" name="sitename" value="<?php print $level; ?>"></td></tr>

</table>

<br />
<input type="submit" name="submit" value="Update">
</form>
<a href="viewedit.php" target="_self">Go back without saving any changes</a>
</body>
</html>

jackpf · April 12, 2009

$db= mysql_connect("localhost", "root") or die(mysql_error());
mysql_select_db("site") or die(mysql_error());

That gave no errors?

You obviously have not done as I told you.

ipodman · April 12, 2009

$db= mysql_connect("localhost", "root") or die(mysql_error());
mysql_select_db("site") or die(mysql_error());
That gave no errors?

You obviously have not done as I told you.

Like i said, No errors which using your code, but I also added the password to ("localhost, "root", "") to ensure access to the database. Also tried the script just suggested above. Still the same problem

jamesxg1 · April 12, 2009

$db= mysql_connect("localhost", "root") or die(mysql_error());
mysql_select_db("site") or die(mysql_error());
That gave no errors?

You obviously have not done as I told you.
Like i said, No errors which using your code, but I also added the password to ("localhost, "root", "") to ensure access to the database

here mate instead of using connection on each page just save this and include it in each page.

<?php

$connection = mysql_connect("DB_SERVER","DB_USER","DB_PASS") 
    or die ("Couldn't connect to server."); 
$db = mysql_select_db("DB_NAME", $connection) 
    or die("Couldn't select database."); 

if(!get_magic_quotes_gpc())
{
  $_GET = array_map('mysql_real_escape_string', $_GET); 
  $_POST = array_map('mysql_real_escape_string', $_POST); 
  $_COOKIE = array_map('mysql_real_escape_string', $_COOKIE);
}
else
{  
   $_GET = array_map('stripslashes', $_GET); 
   $_POST = array_map('stripslashes', $_POST); 
   $_COOKIE = array_map('stripslashes', $_COOKIE);
   $_GET = array_map('mysql_real_escape_string', $_GET); 
   $_POST = array_map('mysql_real_escape_string', $_POST); 
   $_COOKIE = array_map('mysql_real_escape_string', $_COOKIE);
}

?>

ipodman · April 12, 2009

$db= mysql_connect("localhost", "root") or die(mysql_error());
mysql_select_db("site") or die(mysql_error());
That gave no errors?

You obviously have not done as I told you.
Like i said, No errors which using your code, but I also added the password to ("localhost, "root", "") to ensure access to the database

here mate instead of using connection on each page just save this and include it in each page.

<?php

$connection = mysql_connect("DB_SERVER","DB_USER","DB_PASS") 
    or die ("Couldn't connect to server."); 
$db = mysql_select_db("DB_NAME", $connection) 
    or die("Couldn't select database."); 

if(!get_magic_quotes_gpc())
{
  $_GET = array_map('mysql_real_escape_string', $_GET); 
  $_POST = array_map('mysql_real_escape_string', $_POST); 
  $_COOKIE = array_map('mysql_real_escape_string', $_COOKIE);
}
else
{  
   $_GET = array_map('stripslashes', $_GET); 
   $_POST = array_map('stripslashes', $_POST); 
   $_COOKIE = array_map('stripslashes', $_COOKIE);
   $_GET = array_map('mysql_real_escape_string', $_GET); 
   $_POST = array_map('mysql_real_escape_string', $_POST); 
   $_COOKIE = array_map('mysql_real_escape_string', $_COOKIE);
}

?>

Hey,

Thats the way I was going to do it, to call it from config file. But still the problem remains, I have tried doing it with another table and it still doe snot change even when it says it has. And no errors are showing

jamesxg1 · April 12, 2009

$db= mysql_connect("localhost", "root") or die(mysql_error());
mysql_select_db("site") or die(mysql_error());
That gave no errors?

You obviously have not done as I told you.
Like i said, No errors which using your code, but I also added the password to ("localhost, "root", "") to ensure access to the database
here mate instead of using connection on each page just save this and include it in each page.
<?php

$connection = mysql_connect("DB_SERVER","DB_USER","DB_PASS") 
    or die ("Couldn't connect to server."); 
$db = mysql_select_db("DB_NAME", $connection) 
    or die("Couldn't select database."); 

if(!get_magic_quotes_gpc())
{
  $_GET = array_map('mysql_real_escape_string', $_GET); 
  $_POST = array_map('mysql_real_escape_string', $_POST); 
  $_COOKIE = array_map('mysql_real_escape_string', $_COOKIE);
}
else
{  
   $_GET = array_map('stripslashes', $_GET); 
   $_POST = array_map('stripslashes', $_POST); 
   $_COOKIE = array_map('stripslashes', $_COOKIE);
   $_GET = array_map('mysql_real_escape_string', $_GET); 
   $_POST = array_map('mysql_real_escape_string', $_POST); 
   $_COOKIE = array_map('mysql_real_escape_string', $_COOKIE);
}

?>
Hey,

Thats the way I was going to do it, to call it from config file. But still the problem remains, I have tried doing it with another table and it still doe snot change even when it says it has. And no errors are showing

ok put this at the top of your page...

ini_set('display_errors', 1);
error_reporting(E_ALL);

ipodman · April 12, 2009

$db= mysql_connect("localhost", "root") or die(mysql_error());
mysql_select_db("site") or die(mysql_error());
That gave no errors?

You obviously have not done as I told you.
Like i said, No errors which using your code, but I also added the password to ("localhost, "root", "") to ensure access to the database
here mate instead of using connection on each page just save this and include it in each page.
<?php

$connection = mysql_connect("DB_SERVER","DB_USER","DB_PASS") 
    or die ("Couldn't connect to server."); 
$db = mysql_select_db("DB_NAME", $connection) 
    or die("Couldn't select database."); 

if(!get_magic_quotes_gpc())
{
  $_GET = array_map('mysql_real_escape_string', $_GET); 
  $_POST = array_map('mysql_real_escape_string', $_POST); 
  $_COOKIE = array_map('mysql_real_escape_string', $_COOKIE);
}
else
{  
   $_GET = array_map('stripslashes', $_GET); 
   $_POST = array_map('stripslashes', $_POST); 
   $_COOKIE = array_map('stripslashes', $_COOKIE);
   $_GET = array_map('mysql_real_escape_string', $_GET); 
   $_POST = array_map('mysql_real_escape_string', $_POST); 
   $_COOKIE = array_map('mysql_real_escape_string', $_COOKIE);
}

?>
Hey,

Thats the way I was going to do it, to call it from config file. But still the problem remains, I have tried doing it with another table and it still doe snot change even when it says it has. And no errors are showing
ok put this at the top of your page...
ini_set('display_errors', 1);
error_reporting(E_ALL);

Hey,

Now I get errors:-

on update.php

Notice: Undefined index: name in C:\xampplite\htdocs\shop\admin\update.php on line 11

Notice: Undefined index: sale in C:\xampplite\htdocs\shop\admin\update.php on line 12

Notice: Undefined index: cat in C:\xampplite\htdocs\shop\admin\update.php on line 13

But no errors on edit.php

jamesxg1 · April 12, 2009

ok this means they have no value, can i just ask are the $_GET 's getting the info from your edit.php ?

jamesxg1 · April 12, 2009

replace all the $_GET 's you have on update.php to. . .

if (isset($_POST['pro_id']))
$pro_id = mysql_real_escape_string($_POST['pro_id']);
$name =  mysql_real_escape_string($_POST['name']);
$sale =  mysql_real_escape_string($_POST['sale']);
$cat =   mysql_real_escape_string($_POST['cat']);
$id =    mysql_real_escape_string($_POST['id']);
$level = mysql_real_escape_string($_POST['level']);

ipodman · April 12, 2009

replace all the $_GET 's you have on update.php to. . .

if (isset($_POST['pro_id']))
$pro_id = mysql_real_escape_string($_POST['pro_id']);
$name =  mysql_real_escape_string($_POST['name']);
$sale =  mysql_real_escape_string($_POST['sale']);
$cat =   mysql_real_escape_string($_POST['cat']);
$id =    mysql_real_escape_string($_POST['id']);
$level = mysql_real_escape_string($_POST['level']);

With the code you have given above the problem is solved!

jamesxg1 · April 12, 2009

replace all the $_GET 's you have on update.php to. . .

if (isset($_POST['pro_id']))
$pro_id = mysql_real_escape_string($_POST['pro_id']);
$name =  mysql_real_escape_string($_POST['name']);
$sale =  mysql_real_escape_string($_POST['sale']);
$cat =   mysql_real_escape_string($_POST['cat']);
$id =    mysql_real_escape_string($_POST['id']);
$level = mysql_real_escape_string($_POST['level']);

With the code you have given above the problem is solved!

so its done yes ?

ipodman · April 12, 2009

replace all the $_GET 's you have on update.php to. . .

if (isset($_POST['pro_id']))
$pro_id = mysql_real_escape_string($_POST['pro_id']);
$name =  mysql_real_escape_string($_POST['name']);
$sale =  mysql_real_escape_string($_POST['sale']);
$cat =   mysql_real_escape_string($_POST['cat']);
$id =    mysql_real_escape_string($_POST['id']);
$level = mysql_real_escape_string($_POST['level']);

With the code you have given above the problem is solved!

so its done yes ?

Yes done, Thank you very much!!!!

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Updating a mysql record error

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