"Check New Username" function without "Insert

tom fallow · Joined: 30 Dec 2005 Posts: 27 Location: PA

Hello again, have another problem I've been pulling my hair out over (what little I have left)!

I have a multiple form (4 pages). I've used hidden fields to carry the form to the last page, where the information is successfully submitted.

On the first page of the form, I have a textfield called, "UserName". I need to check the entered value against MySQL database and echo and message to the user if the "Username" is already taken.

Because I am not using the "insert record" server behavior in Dreamweaver, I cannot use the "Check New Username" function.

I'm not very familiar with php code yet, although I've been using snippets here and there.

Any help would be greatly appreciated!!

Thanks

Fed · Site Admin Joined: 12 Aug 2005 Posts: 176

this tutorial is a user registration system, in this specific page you find the existing user check http://www.tutorialstream.com/tutorials/php-mysql/create_mysql_table_and_insert_values.php
you find there links to the previous page of the tutorials that might help you to better understand how it all works Wink

basically you run a query for that username, if nothing is returned you can go ahead and create the user

tom fallow · Joined: 30 Dec 2005 Posts: 27 Location: PA

I am well familiar with this tutorial. I've tried setting up the code for the username and querying the database in the fashion described.

1) I end up getting a parse error.

2) It looks like code similar to DMX's "Insert Record" is below. Is the specific code used to query the database dependent on the "Insert into..." code?

3) I have two tables in the database so is there specific code used to select the specific table?

Hope this isn't too confusing.

I can post the code if you like.

Fed · Site Admin Joined: 12 Aug 2005 Posts: 176

well when using 2 tables you should make some joins, this is a pure MySQL issue

what kind of parse error are you getting?

by the way I think that you need only 1 table to check if the username exist

I don't know what you mean with depending on the insert into code, they are 2 separate queries, they obviously use the same $user variable that is checked and eventually added to the database

tom fallow · Joined: 30 Dec 2005 Posts: 27 Location: PA

Well, I'm getting an unexpected $end on line 1034....I don't even have 1034 lines!!

I am working on creating a rent apartment type of website. I have two tables: one for the user registration/login process and another for the properties (so far).

With this specific page I need to check if the "propertyname" exists and if it does, echo an alert "propertyname exists, please choose another" on the same page. This page does not add data to the database, but posts the data to hidden fields on the next two pages, where it is then added to the database.

Here's the <head> part of the code...
Thanks for your input thus far...I've been very frustrated on this...

<?php require_once('Connections/connectdb.php');
$user = $_POST['Propertyname'];//get propertyname from form
$connection = mysql_connect("rentlogin", "owner_prop", "root");
//now we choose the database to be used
@mysql_select_db(owner_prop) or die( "Unable to select database");
//here we declare our sql query statement to see if the user already exist
$check = mysql_query("select propertyname from users where propertyname=\"$user\"");
$returned = mysql_fetch_array($check);
//if a user with the same username is returned we redirect the users to a
//previously created error page
if(!empty($returned))
{
header("Location: error-userexists.php"); //the user will be sent to this page
mysql_close($connection); // and we close the connection to the database
Die();
?>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
<title>Untitled Document</title>
<style type="text/css">

</style>
<script language="JavaScript" type="text/JavaScript">

</script>
</head>

tom fallow · Joined: 30 Dec 2005 Posts: 27 Location: PA

Ok, I've reworked the code for the query. I am not getting any more parse errors. The problem is that I don't get any message if the propertyname does exist. The query is definately successful in connecting to the database. Any help would be most appreciated.

<?php require_once('Connections/connectdb.php');
$connectdb;
@mysql_select_db(rentlogin) or die( "Unable to select database");
$Propertyname = $_POST['Propertyname'];
$check = mysql_query("select Propertyname from owner_prop where Propertyname=\"$Propertyname\"");
$returned = mysql_fetch_array($check);
if (isset($_POST['Submit']))
if(!empty($returned))
{ echo 'Propertyname already exists, please choose another name.';
mysql_close($connectdb); // and we close the connection to the database
Die();
}

?>

tom fallow · Joined: 30 Dec 2005 Posts: 27 Location: PA

I must be missing something as I've reworked the code again and still there is no message being displayed. Help!

<?php require_once('Connections/connectdb.php'); ?>
<?php
if (isset($_POST['submit'])) // if form submitted - process it
$column = "Propertyname";
$table = "owner_prop";
$sql = "SELECT $column FROM $table WHERE Propertyname = '{$_POST['Propertyname']}'";
$result = @mysql_query($sql);
$row = @mysql_fetch_object($result);
if ($row) // if propertyname already taken
{
echo 'Sorry, the propertyname you specified is already taken. Please choose another.';
mysql_close(connectdb); // close database connection
}

Fed · Site Admin Joined: 12 Aug 2005 Posts: 176

tom fallow · Joined: 30 Dec 2005 Posts: 27 Location: PA

Thanks for the input.

I decided to work with the code in a new page with a textfield (Propertyname) and a submit button only. With this code, I get an error that reads, "query 1: unable to select database". From my understanding of the code, my database (rentlogin) is selected and my table (owner_prop) is selected as well. What am I doing wrong here?

<?php require_once('Connections/connectdb.php');
if (isset($_POST['Submit'])) // if form submitted - process it
mysql_select_db(rentlogin) or die( "Unable to select database");
$Propertyname = $_POST['Propertyname'];
$column = "Propertyname";
$table = "owner_prop";
$sql = mysql_query("SELECT 'Propertyname' FROM $table WHERE 'Propertyname' = '$Propertyname'") or die ("query 1: " . mysql_error());
$mysql_num = mysql_num_rows($sql);
if ($mysql_num >= 1)
{
echo 'Cannot insert duplicates';
}
?>

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
<title>Untitled Document</title>
</head>
<body>
<form action="" method="post" name="Propertyname" id="Propertyname">
<p>
<input type="text" name="textfield">
</p>
<p>
<input type="submit" name="Submit" value="Submit">
</p>
</form>
</body>
</html>

Fed · Site Admin Joined: 12 Aug 2005 Posts: 176

this error means that your script can't connect to the database

I think that the database name must necessarily be passed as a string so I'd use: @mysql_select_db("rentlogin") or die( "Unable to select database");

or @mysql_select_db($variable) or die( "Unable to select database"); but this $variable should be a string like "dbname"

I hope I understood the error Smile

tom fallow · Joined: 30 Dec 2005 Posts: 27 Location: PA

Thanks Fed, I made the changes as you suggested. However, nothing changed as far as the error.

I made a mistake...the error reads, "query 1: no database selected". I think the database is selected, at least in the line that reads:

@mysql_select_db("rentlogin") or die("Unable to select database");

...or I would get an error that reads, "unable to select database" instead of, "no database selected".

I do have two tables in the database, but I don't think that's the problem because the proper table is selected on these lines:
$table = "owner_prop";
$sql = mysql_query("SELECT 'Propertyname' FROM $table WHERE 'Propertyname' = '$Propertyname'") or die ("query 1: " . mysql_error());

Still not sure why the error "no database selected".

Fed · Site Admin Joined: 12 Aug 2005 Posts: 176

well I assume that you are connecting to your database into the file Connections/connectdb.php

are you sure your connection details are right? host, databasename, user, password?

try having a look here if you doubt your code is connecting http://www.tutorialstream.com/tutorials/php-mysql/php_connect_mysqldb.php

usually when someone can't select a db this is the problem Very Happy

tom fallow · Joined: 30 Dec 2005 Posts: 27 Location: PA

Thanks Fed,
Making adjustments to connecting to the database seemed to do the trick. However, using this code, I don't get the echo expression when the textfield matches that of the record in the database. Do I need to have a return command here?

<?php
require_once('db1config.php');
require_once('db1connect.php');
if (isset($_POST['Submit'])) // if form submitted - process it
@mysql_select_db("rentlogin") or die("Unable to select database");
$Propertyname = $_POST['Propertyname'];
$column = "Propertyname";
$table = "owner_prop";
$sql = mysql_query("SELECT $column FROM $table WHERE 'Propertyname' = '$Propertyname'") or die ("query 1: " . mysql_error());
$mysql_num = mysql_num_rows($sql);
if ($mysql_num >= 1)
{
echo 'Cannot insert duplicates';
}
?>

Fed · Site Admin Joined: 12 Aug 2005 Posts: 176

have you tried to verify the value of $mysql_num ?
if the query was wrong (assuming that you have populated the db with that username I hope Very Happy

) it wouldn't be >= 1 so it wouldn't execute the if section

I tend to use echo with double quotes when there are no variables inside, but I think it should work as well with the single quote Neutral

tom fallow · Joined: 30 Dec 2005 Posts: 27 Location: PA

I figured I'd add a redirect to the "if" clause of the code, and the redirect does not function when the Propertyname field exists. Any ideas?

Also, I'm not inserting these values, but posting them in hidden fields to subsequent pages, where the past page does the inserting (not sure if this affects the process).

<?php
require_once('db1config.php');
require_once('db1connect.php');
if (isset($_POST['Submit'])) // if form submitted - process it
@mysql_select_db("rentlogin") or die("Unable to select database");
$Propertyname = $_POST['Propertyname'];
$column = "Propertyname";
$table = "owner_prop";
$sql = mysql_query("SELECT $column FROM $table WHERE 'Propertyname' = '$Propertyname'") or die ("query 1: " . mysql_error());
$mysql_num = mysql_num_rows($sql);
if ($mysql_num >= 1)
{
header("Location: error-userexists.php");
echo "Cannot insert duplicates";
Die();
}
?>