[RESOLVED] Two submit buttons in one form ....

anoopd

Im facing a problem with a preview page having two submit buttons just like our "New Thread Post" . One is an "Edit" button and another a "Save" button which saves all fields to db.The problem is when i click "Save" an error is showing up saying

Undefined index: edit in C:\Inetpub\wwwroot\preview1.php on line 9

This edit is the value of edit button for which i want to go back to the page with form .Please somebody help me to solve this out .I would also like to know how to go back to the edit page so that the user can edit the contents if he needs.I am a newbie and this is my first project.Please help

macmonkey

I'd have to see your form code and a little of the php to help ya out.

frost110

Post your code so we can look at it and help you.

--FrosT

jazz_snob

Are you checking $POST['edit']? (or $GET['edit'], same result) Try putting code in if( isset( $_POST['edit'])){//do the edit thang}

anoopd

That solved my Problem ....Thanks a lot jazz !!🙂.I changed it to "isset" and now it is working fine!!Ty so much and ty phpbuilder.I would like to get more help regarding going back to my edit page.I need all my data to be there in same text field when i press "Edit" button in preview form.How can i achive that .Im using "header(Location:quote.html) on the "Edit" though it takes me to the quote page the data is not there in textarea.Please help .The code is

quote.html
<html>
<head><title>quote</title></head>
<body>
<form action=preview.php method=post>
Address:<textarea name=address rows=3></textarea><br>
<INPUT type="submit" value="Preview"/>
<INPUT type=reset value="Clear"/>
</form>
</body>
</html>

preview.php
<?php
$addr=$_POST['address'];
echo'<form action=preview1.php method=post>';
echo $addr,'<input type=hidden name=address2 value="'.$addr.'">';
echo'<INPUT type="submit" value="Edit"/ name=edit><INPUT type="submit" value="Save"/ name=save>';
echo'</form>';
?>

preview1.php
<?php
$addr1=$_POST['address2'];

if(isset($_POST['edit'])) {
header('Location: quote.html');
}

elseif(isset($_POST['save'])) {

$host = "localhost";
$user = "username";
$pass = "password";
$dbname = "dbname";

$conn = mysql_connect($host,$user,$pass) or die (mysql_errno().": ".mysql_error()."<BR>");
mysql_select_db($dbname);

$request = "INSERT INTO tablename (address)value('".$addr1."')";

$results = mysql_query($request,$conn);

if($results)
{
echo "Document SAVED<br>";

}
else
{
echo "There was an error. Try Again.";
}

}

Here is there any method for storing the input in three lines(The address) as it is
entered and to display it . Hope my problem will be solved out with so many seniors like you being here.

                                    Thanking you
                                                           anoopd