I want to use a php form textfield to change a value in my databas. I got this far by im keep getting this error, please help me figure out whats wrong.

PHP Code:

<?php

 $server = "localhost";
  $username = "admin";
  $password = "admin";
  $database = "local_db";
  $con = mysql_connect($server, $username, $password);
  $ok = mysql_select_db($database, $con);
      
        $Cname = mysql_real_escape_string($_POST['newname']);

    $sql = "UPDATE people SET name='{$Cname}' WHERE id=1";
    mysql_query($sql) or die(mysql_error());

        

 
?>


<form method="post" action="">
  
    <input name="newname" type="text" id="textfield"/>
    
    <input type="submit" name="change" value="Change now" />
    
</form>

Notice: Undefined index: newname in this line,

PHP Code:

$Cname = mysql_real_escape_string($_POST['newname']); 


it's probably only when you load this page without submitting it.
If you submit the form, the result should be ok, I don't see any errors in your code.

The fact is, that php will run through your update code at each time the page is loaded, not only when a post is made.
It's your job to tell php when to do the update.

PHP Code:

<?php

  $server = "localhost";
  $username = "admin";
  $password = "admin";
  $database = "local_db";
  $con = mysql_connect($server, $username, $password);
  $ok = mysql_select_db($database, $con);
  
  if(sizeof($POST)>0){
    //we go in here only where there was a POST submitted
    $Cname = mysql_real_escape_string($_POST['newname']);
    $sql = "UPDATE people SET name='{$Cname}' WHERE id=1";
    mysql_query($sql) or die(mysql_error());
  }
?>
<form method="post" action="">
    <input name="newname" type="text" id="textfield"/>
    <input type="submit" name="change" value="Change now" />
</form>

You could also use:

PHP Code:

<?php

  $server = "localhost";
  $username = "admin";
  $password = "admin";
  $database = "local_db";
  $con = mysql_connect($server, $username, $password);
  mysql_select_db($database, $con); // Don't actually need to make this a variable. 
  
  if($_POST['change'] === 'Change now'){ // If the form has been submitted.
    $sql = "UPDATE people SET name='".mysql_real_escape_string($_POST['newname'])."' WHERE id=1 LIMIT 0,1"; // Changed this line also.
    mysql_query($sql) or die(mysql_error());
echo 'Updated.';
  }
?>
<form method="post" action="">
    <input name="newname" type="text" id="textfield"/>
    <input type="submit" name="change" value="Change now" />
</form>

* Edit: Fixed that error.

Getting the following error from both above codes Notice: Undefined variable: POST

$_POST

Just to confirm that, and so you know exactly where:

PHP Code:


<?php

  $server = "localhost";
  $username = "admin";
  $password = "admin";
  $database = "local_db";
  $con = mysql_connect($server, $username, $password);
  mysql_select_db($database, $con); // Don't actually need to make this a variable. 
  
  if($_POST['change'] === 'Change now'){ // If the form has been submitted.
    $sql = "UPDATE people SET name='".mysql_real_escape_string($_POST['newname'])."' WHERE id=1 LIMIT 0,1"; // Changed this line also.
    mysql_query($sql) or die(mysql_error());
echo 'Updated.';
  }
?>
<form method="post" action="">
    <input name="newname" type="text" id="textfield"/>
    <input type="submit" name="change" value="Change now" />
</form>

</span>

you can also use the following to execute the query only when the form is submitted

PHP Code:

<?php $server = "localhost";
  $username = "admin";
  $password = "admin";
  $database = "local_db";
  $con = mysql_connect($server, $username, $password);
  mysql_select_db($database, $con); // Don't actually need to make this a variable. 
  
  if($_POST){ // since there a single form that can be submitted
    $sql = "UPDATE people SET name='".mysql_real_escape_string($_POST['newname'])."' WHERE id=1 LIMIT 0,1"; // Changed this line also.
    mysql_query($sql) or die(mysql_error());
echo 'Updated.';
  }
?>
<form method="post" action="">
    <input name="newname" type="text" id="textfield"/>
    <input type="submit" name="change" value="Change now" />
</form>

Thank you guys!!

This site is great. After trying things for two days, Tripy's solution worked for me. Now I have to build on that to enable mysql updating from a forum with multiple records and a submit button.

Thanks so much .. this was the first time I could update the database from a form entry.

Just a thught.
If this is the only action you take involving the database, that is if you don't do other things too with the database, you can also put this

PHP Code:

  $server = "localhost";
  $username = "admin";
  $password = "admin";
  $database = "local_db";
  $con = mysql_connect($server, $username, $password);
  $ok = mysql_select_db($database, $con); 


in the if statement. It's unnecessary to open a database connection if your not going to use it.