PHP Forum

**sith717** · 11-04-2009, 07:43 PM

How do I do it so when this page gets run by itself it doesn't update the table, since if it does it will just make all the fields blank.

PHP Code:


			
<?php
require_once ('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());
?>

Instead make it update the table update only when a form updates it?

**orionoreo** · 11-04-2009, 07:58 PM

side question... how come your sig says php and web design professionally done... pm for html, php help when you're asking a lot of basic information

don't take it offensively I'm just curious

PHP Code:


			
<?php

<?php
require_once ('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;  

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

**chrishirst** · 11-04-2009, 08:00 PM

PHP Code:


			
if($_POST['submit_button_name'] != "" ) {

    do update; 

} else {

    do not update;

}

**sith717** · 11-04-2009, 08:04 PM

Thanks I used something a little more simple now I have another question...

I am trying to do it so the value in the text box is the current value that is in the database so I do not have to retype, and instead edit it or add on to it.

PHP Code:


			
<html>
<body>
<?php
if (isset ($_POST['submit'])) // if the form was submitted, display their name
{
require_once ('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());
echo "hello";
}
//form hasent been submitted
{

?>


<form action="<?php echo $_SERVER['PHP_SELF'] ?>" method="post">
Firstname: <input type="text" value="<?php echo $row['FirstName'];?>" name="firstname" /><br>
Lastname: <input type="text" value="<?php echo $row['LastName'];?>" name="lastname" /><br>
middlename: <input type="text" value="<?php echo $row['MiddleName'];?>" name="middlename" /><br>
<input type="submit" id="submit" name="submit" value="Submit!">
</form>
<?php
}
?>


</body>
</html>

Basicly the value isnt showing up.

PHP Code:


			
<input type="text" value="<?php echo $row['MiddleName'];?>" name="middlename" />

How do I fix that?

**orionoreo** · 11-04-2009, 08:33 PM

you didn't even call $row yet

PHP Code:


			
$row = mysql_fetch_array(mysql_query("SELECT * FROM 'testing' WHERE id='$id'"));

and btw your method, somebody can still set text fields to null and empty out your columns

**sith717** · 11-04-2009, 08:54 PM

Thanks here is my final code:

PHP Code:


			
<html>
<body>
<?php
if (isset ($_POST['submit'])) // if the form was submitted, display their name
{
require_once ('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());
echo "hello";
}

require_once ('inc/config.php');
$query = "SELECT * FROM testing";
$result = mysql_query($query) or die(mysql_error());
while ($row = mysql_fetch_array ($result))
{
?>



<form action="<?php echo $_SERVER['PHP_SELF'] ?>" method="post">
Firstname: <input type="text" value="<?php echo $row['FirstName'];?>" name="firstname" /><br>
Lastname: <input type="text" value="<?php echo $row['LastName'];?>" name="lastname" /><br>
middlename: <input type="text" value="<?php echo $row['MiddleName'];?>" name="middlename" /><br>
<input type="submit" id="submit" name="submit" value="Submit!">
</form>
<?php
}
?>

</body>
</html>

Now I have a new problem:

I want to echo what is currently in the database but it wont show up

PHP Code:


			
<?php
// show errors if any
error_reporting(E_ALL);
ini_set('display_errors', '1');

// require a file
require_once ('inc/config.php');

// select row from table
$query = "SELECT * FROM testing";

// check if is valid if it is then show results if not then die
$result = mysql_query($query) or die(mysql_error()); 
while ($row = mysql_fetch_array ($result)) 
{
?>
First Name:
<?php $row['FirstName']; ?>
<br>Last Name:
<?php $row['LastName']; ?>
<br>Middle Name:
<?php $row['MiddleName']; ?>
<?
}
?>

**lizciz** · 11-05-2009, 02:49 AM

You forgot to actually print it.

PHP Code:


			
<?php $row['Firstname']; ?>

should be

PHP Code:


			
<?php echo $row['Firstname']; ?>

**Envision_frodo** · 11-05-2009, 01:00 PM

Shorthand:

PHP Code:


			
<?= $row[0] ?>

Note: "<?=" != "<?php"

**chrishirst** · 11-05-2009, 04:08 PM

depends on how the server is set up.

**Envision_frodo** · 11-05-2009, 05:57 PM

Sorry, should've mentioned... In order to use the shorthand method you must have the "short_open_tag" option enabled in your php.ini file. Also, the use of shorthand is highly discouraged given the decreased portability.

That doesn't mean you can't still love it. As a pure text-only coder it just "feels" cleaner to me.