This form is not working.

When I click Save it shows this error.

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'add SET (`id`, `code`) VALUES (NULL, 'sdfsd')' at line 1

This is the code.

PHP Code:

<form method="post">
<?php
// if
if(isset($_POST['preview'])) {
// show code as html
echo $_POST['code']."<br />";
// show save and edit buttons
echo "<input type='submit' name='save' id='submit' value='Save' /><input type='submit' name='edit' id='submit' value='Edit code' />";
}
else { // if not preview then show form
?>
    <label>Enter Add Code:</label>
    <textarea name="code" style="width: 200px; height: 100px;"><?php
                                                                // if edit then show code in textarea
                                                                if(isset($_POST['edit'])) {
                                                                    echo $_POST['preview']; 
                                                                } // if not edit
                                                                ?></textarea>
    <input type="submit" name="submit" id="submit" value="Save" />
    <input type="submit" name="preview" id="submit" value="Preview" />
</form>
<?php
// check if submited code
if(isset($_POST['submit'])) {

/* connect to the database */
// conect to the server
$connect = mysql_connect("localhost","db_username","db_password");
    // if connected to the server continue
    if($connect == TRUE) {
        // Select the database and check if connected
        if(mysql_select_db("db_name") != TRUE) {
            // if not connected display error
            exit("<span style='color: red'>Can't connect to the MySQL database. Please contact the webmaster.</body></html>");
        // end if can not connect to database
        }
    // if not connected to the server then contiunue
    } else{
        // display error if not connected to the server
        exit("<span style='color: red'>Can't connect to the MySQL server. Please contact the webmaster.</body></html>");
    // end if not connected to the server
    }

// insert add into database
$sql = mysql_query("UPDATE add SET (`id`, `code`) VALUES (NULL, '".$_POST['code']."')") or die (mysql_error());

if($sql = TRUE) {
// echo complete msg or divert
echo "the code has been updated";
// divert back to this page
header('location: forms.php');
} // end complete
} // end submit
} // end if not preview.
?>

---

More about the form:

Basicly what I need, is a Ad Managment script.

Some things In it:

A form, basicly a textarea where I can insert the code and click update.

When I click Update, on the index it will show the Banner/Advert I have placed.

Like It would connect to a database or something.

Thats all I need, no counters, nothing special, a form thats it.

The first person being able to help me out.

----

• I would create a page in the admin section that displays the current ad with a form below.
• The form would be a big text area that updates the database, the value of the form would come from the database displaying the html code in the text area.
• Once the form is edited and submitted the page would reload displaying the new ad.
• Each ad would require its own page, re-using the original code.
• Maybe create a preview button to make sure you entered the correct code before you update the database.

I will be able to do something in return, if that means, give you FREE hosting with any features you want or designing you a logo.

Bump. I still need help.

can you show us your create table statement for the table?

I dont think id can't be null...unless it's not primary key

Here

PHP Code:

<?php
/* connect to the database */
// conect to the server
$connect = mysql_connect("localhost","username","password");
    // if connected to the server continue
    if($connect == TRUE) {
        // Select the database and check if connected
        if(mysql_select_db("db_name") != TRUE) {
            // if not connected display error
            exit("<span style='color: red'>Can't connect to the MySQL database. Please contact the webmaster.</body></html>");
        // end if can not connect to database
        }
    // if not connected to the server then contiunue
    } else {
        // display error if not connected to the server
        exit("<span style='color: red'>Can't connect to the MySQL server. Please contact the webmaster.</body></html>");
    // end if not connected to the server
    }
?>

<form method="post">
<?php
// if preview
if(isset($_POST['preview'])) {
// show code as html
$code = stripslashes($_POST['code']);
echo $code."<br />";
echo "<input type='hidden' name='code' value='".$code."' / >";
// show save and edit buttons
echo "<input type='submit' name='submit' id='submit' value='Save' /><input type='submit' name='edit' id='submit' value='Edit code' />";

} else { // if not preview then show form
?>
    <label>Enter Add Code:</label>
    <textarea name="code" style="width: 200px; height: 100px;"><?php
                                                                /* if edit then show code in textarea */
                                                                if(isset($_POST['edit'])) {
                                                                    echo stripslashes($_POST['code']); 
                                                                } // if not edit
                                                                ?></textarea>
    <input type="submit" name="submit" id="submit" value="Save" />
    <input type="submit" name="preview" id="submit" value="Preview" />
</form>
<?php
// check if submited code
if(isset($_POST['submit'])) {

// check the database
$sql = mysql_query("SELECT * FROM `addvert`") or die (mysql_error());
// count results returned
$count = mysql_num_rows($sql);
// check wether to update or insert.
$code = mysql_real_escape_string($_POST['code']);
if($count == 0) {
$add = mysql_query("INSERT INTO `addvert` (`id`, `code`) VALUES (NULL, '$code')") or die (mysql_error());
} else {
$add = mysql_query("UPDATE `addvert` SET code='$code' WHERE id='1'") or die (mysql_error());
} // ends update or insert

if($add == TRUE) {
// echo complete msg or divert
echo "the code has been updated";
/* 
############### You can echo the code above OR refresh the page ###############
divert back to this page

header('location: thispage.php');
*/
} else { // echo error
echo "there was an error with the querys";
} // end complete
} // end submit
} // end if not preview.
?>

To create the databse use:

PHP Code:

CREATE TABLE IF NOT EXISTS `addvert` (
  `id` int(255) NOT NULL auto_increment,
  `code` longtext NOT NULL,
  PRIMARY KEY  (`id`)
) ENGINE=MyISAM  DEFAULT CHARSET=latin1 AUTO_INCREMENT=1 ; 


and to display the code around the website use:

PHP Code:

<?php
/* Make sure that it has the conection scrip above this code on all pages :) */
$sql = mysql_query("SELECT * FROM addvert") or die (mysql_error());
$rows = mysql_fetch_array($sql);
echo stripslashes($rows['code']);
?>

I just need help with:
HOW TO SHOW MULTIPLE ADVERTS IN DIFFERENT PLACES, AND SHOW THE CURRENT DATABASE TEXT IN THE TEXTBOX VALUE.

so you're not getting the error msg like your first post?

PHP Code:

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'add SET (`id`, `code`) VALUES (NULL, 'sdfsd')' at line 1 


to solve this...change your insert statement to be...

PHP Code:

$add = mysql_query("INSERT INTO `addvert` (`code`) VALUES ('$code')") or die (mysql_error()); 


you dont need to define the 'id' column, as it is auto increment, and will take care of itself

and I'm a little confused on exactly what your asking...about adverts in different places.

do you want like an admin page that displays current adverts in the database that shows each specific adverts information?

I have fix it. Thank You anyways.