[RESOLVED] Specify Insert Table On Different Page?

pearjam

So I've got two pages, a Form Page and a Thank You page.

I'm trying to set it up where the Form Page handles:
The form.
The sql insert statement specifying the table to insert into and the fields.
ie:

$sql = "INSERT INTO `tablename` (name, comment) VALUES('{$_POST['name']}', '{$_POST['comment']}')";

...and the Thank You page handles:
Opens mysql db connection.
Runs the insert statement from the Form Page.
Closes db connection.
Says 'thank you'.
Redirects to previous page.

I've got the code for everything else figured out, but I can't figure out how to put the "insert into tablename" code on the Form Page and have it pass the info to the Thank You page.

Any ideas how to do this?

bike5

Why not just run the sql on the processing page then forward to the thankyou page? Look into sessions if you really want to process that on the thankyou page.

pearjam

That's what I'm thinking...

If I do it that way, how can I use one db connect to serve the two sections of the page?

ie: On a page with a form:
one mysql db connect will allow both the table insert statement, and the display from * tablename

bike5

if you do any database stuff on different pages each page will require a new connection. Thsi si where putting DB vars into an include fle and have a database class comes in handy.

pearjam

...how does one put "DB vars into an include file and have a database class?

🙂

If I were to have the mysql connection on the Form Page, how would that work?

I have two areas of the page that need to use the connection (insert contents into a table, and display contents of a table).

Isn't there a way to have the mysql connect info at the top...

display the table contents once then end...

If the form is submitted, write the contents to the table (if not, do nothing).

Here's what I have, and it doesn't work right:

<?php
$connect = mysql_connect("localhost","root","xxxxx") or die(mysql_error());
mysql_select_db("feedback") or die(mysql_error());
$result = mysql_query("SELECT * FROM main");
while($row = mysql_fetch_array($result))
  {
  echo $row['time'] . "&nbsp;" . $row['name'] . "&nbsp;" . $row['site'] . "<br /><i>" . $row['comment'] . "</i><br /><br />";
  }
mysql_close($connect);
?>

<div id="right">
<form action="process.php" method="post">
<input type="text" name="name" size="20" value="your name" style="border:0px;color:#A7A7A7;font-size:11px;" onFocus="this.value=''" /><br /><br />
<input type="text" name="site" size="20" value="your website" style="border:0px;color:#A7A7A7;font-size:11px;" onFocus="this.value=''" /><br /><br />
<textarea name="comment" rows="12" cols="19" style="border:0px;color:#A7A7A7;font-size:11px;" onFocus="this.value=''">your feedback</textarea><br /><br />
<input type="submit" value="post" />
</form>

<?php
$connect = mysql_connect("localhost","root","xxxxx") or die(mysql_error());
mysql_select_db("feedback") or die(mysql_error());
$_POST = array_map('mysql_real_escape_string', $_POST);
$time = date("[d My]");
$site = $_POST["site"];
$sql = "INSERT INTO `main` (time, site, name, comment) VALUES('$time', '{$_POST['site']}', '{$_POST['name']}', '{$_POST['comment']}')";
mysql_query($sql) or die(updating);
mysql_close($connect);
?>

bike5

Instead of hard coding this

$connect = mysql_connect("localhost","root","xxxxx") or die(mysql_error());
mysql_select_db("feedback") or die(mysql_error());

you could do

include_once('includes/db.php'); // put the database variables into this file
$connect = mysql_connect($dbhost,$dbusername,$dbuserpass) or die(mysql_error());
mysql_select_db($dbname) or die(mysql_error());

As far as using a database class, you might just want to search google for" php mysql database class" and see what shows up.

cahva

You dont have to open the the connection everytime you do a query. Just once before using mysql is enough. You dont have to close the connection. When script exection ends, it will close the connection automatically.

Bjom

First of all be aware that each form is a single request and that each request is completely unrelated to any other request. There is no persistence whatsoever. PHP forgets everything that has happened before.

Do you actually need to do any DB stuff on the "form"?

As I see it you just need to handle everything on the thank you page and that's where you put your DB stuff.

I would suggest you use mysqli instead of mysql (and maybe even use it OO style). And I would also suggest that or die must die.

Bjom

pearjam

cahva;10918438 wrote:
You dont have to open the the connection everytime you do a query. Just once before using mysql is enough.

What do you mean?

I'm looking to keep the code optimized and streamlined, but I've got no real idea of what I'm doing.... I also need to add protection to the form, but I'm having a hard time with fitting examples on the web into the code I have.

I can show you all what I have so far - and I could use any ideas on how to slim it down and add form protection (like strip tags etc...).

<?php
$connect = mysql_connect("localhost","root","xxxxxx") or die(mysql_error());
mysql_select_db("feedback") or die(mysql_error());
$_POST = array_map('mysql_real_escape_string', $_POST);
$time = date("[d My]");
$sql = "INSERT INTO `{$_POST['table']}` (time, site, name, comment) VALUES('$time', '{$_POST['site']}', '{$_POST['name']}', '{$_POST['comment']}')";
mysql_query($sql) or die(did not update);
$ref = $_SERVER['HTTP_REFERER'];
header( 'refresh: 01; url='.$ref);
mysql_close($connect);
?>

It's basically a mini-feedback form for different pages of my site - so the db is "feedback" and the tables are the page names = different page, different data.