PHP Forum

**sith717** · 02:27 PM

I would like to make this script so that instead of inserting info it updates the code to update the first row.

I thought of adding this to the update

PHP Code:


			
WHERE id='$id'

and having this to say what row

PHP Code:


			
$id = $rows['id'];

So something like this:

PHP Code:


			
<?php
 require('inc/config.php'); 
$id = $rows['id'];
$sql="UPDATE Persons WHERE id='$id' (FirstName, LastName, Age)
VALUES
('$_POST[firstname]','$_POST[lastname]','$_POST[age]')";

if (!mysql_query($sql))
  {
  die('Error: ' . mysql_error());
  }
echo "1 record added";
?>

And the form:

PHP Code:


			
<html>
<body>

<form action="insert.php" method="post">
Firstname: <input type="text" name="firstname" />
Lastname: <input type="text" name="lastname" />
Age: <input type="text" name="age" />
<input type="submit" />
</form>

</body>
</html>

Is that correct?

and mostly will it work?

**chrishirst** · 11-04-2009, 02:33 PM

No.

Update uses the

Code:

UPDATE table SET columnA = valueA, SET columnB = valueB WHERE criteria = value;

syntax.

**sith717** · 02:40 PM

So how would the updated code look like?

I really dont understand.

Like edited so it will work with my form.

**sith717** · 02:51 PM

So something like this?

PHP Code:


			
<?php
include ('inc/config.php');
$id = $rows['id'];  

mysql_query("UPDATE Persons WHERE id='$id' SET Age = '$_POST[age]'
WHERE FirstName = '$_POST[firstname]' AND LastName = '$_POST[lastname]'");

?>

Something like that?

**orionoreo** · 11-04-2009, 03:00 PM

you don't need to do that since your id is unique all you need is

PHP Code:


			
mysql_query("UPDATE Persons SET Age = '$_POST[age]' WHERE id='$id'");

as you said in your last thread about sanitizing input you should keep to that yourself... personally I would not send inputs straight to my db with out verifying the data... in this case you should

PHP Code:


			
if (is_numeric($_POST['aga'])) mysql_query("UPDATE Persons SET Age = '$_POST[age]' WHERE id='$id')");
else echo 'Please enter valid age';

**sith717** · 11-04-2009, 03:04 PM

The age was an example.

So ye. :P

I will have no ages in it.

So it will be like this?

PHP Code:


			
<?php
include ('inc/config.php');
$id = $rows['id'];  

mysql_query("UPDATE Persons WHERE id='$id' SET
FirstName = '$_POST[firstname]', 
LastName = '$_POST[lastname]',
MiddleName = '$_POST[middlename]'
");

?>

Like that? Saying that middle name is a textbox in the form.

**Envision_frodo** · 11-04-2009, 03:12 PM

Better: "...firstname = mysql_real_escape_string(trim(".$_POST['firstname']."))..."

**sith717** · 03:17 PM

Okay I have done that, will it work?

PHP Code:


			
 <?php
include ('inc/config.php');
$id = $rows['id'];  

mysql_query("UPDATE Persons SET
$FirstName = mysql_real_escape_string ($_POST['firstname']),
$LastName = mysql_real_escape_string ($_POST['lastname']),
$MiddleName= mysql_real_escape_string ($_POST['middlename']);
WHERE id='$id'
");
?>

Is that correct? and will it work?

**orionoreo** · 11-04-2009, 03:26 PM

that should work no problem but if it doesn't it maybe the placement of quotations

PHP Code:


			
mysql_query("UPDATE Persons SET
$FirstName = 'mysql_real_escape_string ($_POST[firstname])',
$LastName = 'mysql_real_escape_string ($_POST[lastname])',
$MiddleName= 'mysql_real_escape_string ($_POST[middlename])';
WHERE id='$id'
");

**Envision_frodo** · 11-04-2009, 03:33 PM

If everything else is gravy it should. Note: You will need to have a viable connection open to the database of choice. For more:

Connect - http://www.php.net/manual/en/function.mysql-connect.php
Select - http://www.php.net/manual/en/functio...-select-db.php
Query - http://php.net/manual/en/function.mysql-query.php

Do Something (I suggest: mysql_fetch_row())

Close - http://www.php.net/manual/en/function.mysql-close.php

**sith717** · 11-04-2009, 03:44 PM

Okay I currently got this for the form called page1.php

PHP Code:


			
<html>
<body>

<form action="upage2.php" method="post">
Firstname: <input type="text" name="firstname" />
Lastname: <input type="text" name="lastname" />
middlename: <input type="text" name="middlename" />
<input type="submit" />
</form>

</body>
</html>

and this for the page2.php

PHP Code:


			
<?php
include ('inc/config.php');
$id = $rows['id'];  

mysql_query("
UPDATE Persons SET
$FirstName = 'mysql_real_escape_string ($_POST[firstname])',
$LastName = 'mysql_real_escape_string ($_POST[lastname])',
$MiddleName= 'mysql_real_escape_string ($_POST[middlename])';
WHERE id='$id'
");  

?>

It isnt working at all and I am getting no errors

**sith717** · 11-04-2009, 03:50 PM

I have also tried it like this

PHP Code:


			
<?php
include ('inc/config.php');
$id = $rows['id'];  

$sql = mysql_query ("
   UPDATE
      `persons`
   SET 
      `FirstName` = '".$firstname."', 
      `LastName` = '".$lastname."',
      `LastName` = '".$middlename."' 
   WHERE
      `id` = '".$id."'
"); 

?>

**sith717** · 11-04-2009, 03:52 PM

Im getting these errors:

Quote:

Notice: Undefined variable: rows in /home/runehost/public_html/update.php on line 7

Notice: Undefined index: firstname in /home/runehost/public_html/update.php on line 13

Notice: Undefined index: lastname in /home/runehost/public_html/update.php on line 14

Notice: Undefined index: middlename in /home/runehost/public_html/update.php on line 15

**lizciz** · 04:38 PM

PHP Code:


			
$id = $rows['id'];

What is $rows? You need to have the id of the row that you want to update.
Then do this.

PHP Code:


			
$first  = mysql_real_escape_string($_POST['first']);
$last  = mysql_real_escape_string($_POST['last']);
$middle  = mysql_real_escape_string($_POST['middle']);

// I split the line so it fits in the forum code box
$query = "UPDATE persons SET
      first='$first', last='$last', middle='$middle' WHERE id='$id'";
mysql_query($query) or die(mysql_error());

**sith717** · 11-04-2009, 04:42 PM

PHP Code:


			
<?php
include ('inc/config.php');
$firstname = mysql_real_escape_string ($_POST['firstname']);
$lastname = mysql_real_escape_string ($_POST['lastname']);
$middlename= mysql_real_escape_string ($_POST['middlename']);

$id = 2;  
$sql = mysql_query ("
   UPDATE
      `testing`
   SET 
      `FirstName` = '".$firstname."', 
      `LastName` = '".$lastname."',
      `MiddleName` = '".$middlename."' 
   WHERE
      `id` = '".$id."'
")
OR die (mysql_error());
?>

Fixed so there's no error

Right now it is editing row 2. But in my column I only have 1 row and will have only 1 row, so how would I do it so it automatically edits the only row in the table.

Do I use the * thing?

**lizciz** · 11-04-2009, 06:12 PM

If you know that at all times there will be only one row, you can just skip the WHERE clause, thereby applying the changes to all rows (in this case only one). So:

Code:

"UPDATE testing SET FirstName='$firstname', LastName='$lastname', MiddleName='$middlename'"

**sith717** · 11-04-2009, 06:41 PM

Please do not spam my thread.