need help with modifying a database

not_tekky_ever

02-09-2007, 01:21 PM

Hey folks
I'm new here & very VERY new to PHP & MySQL.

here's the wrap, i need to create a database for calculating somebody's wages.

I've made the calculator, i've created a table that pulls out an employee's information, & i've managed to creat a form & handler that will insert new employee data. However, I dont have a clue how to create a modify page so the user can access the table from the web browser & alter information.

Here's what i've got:

<html>
<head>
</head>
<body>

<?php
$host="local";
$user="c301";
$password="fo";
$dbname="c301";
$tablename="employee";

$link = mysql_connect('localhost', 'mysql_user', 'mysql_password');
if (!$link) {
die('Not connected : ' . mysql_error());
{
db_selected = mysql_select_db($dbname, $link);
if (!$db_selected) {
die ('Can\'t use foo : ' . mysql_error());

$dbname="c3018192";
$tablename="employee";

$query = " UPDATE $tablename SET (`number` , `name` , `address` , `postcode`, `telephone` , `email` , `hours`)

VALUES ('$number', '$name', '$address', '$postcode', '$telephone', '$email', '$hours') where primaryky = `number`;";

if (mysql_select_db($dbname, $link, )){;

print"\"ok\"";
}
else
{
print "\"die('Could not connect: ' . mysql_error())\"";
}
mysql_close($con);
?>

</body>
</html>
[/COLOR]

All I seem to get is:

Parse error: parse error, unexpected '=' in c:\inetpub\wwwroot\04\c301\Advanced Web Applications\test folder\test.php on line 17

Dont suppose anybody can help me? my assignments a week late & the tutor hasn't got any information on his site :confused:

Rob

guelphdad

02-09-2007, 01:54 PM

check the line the error is on:
db_selected
it should most likely be a variable.

not_tekky_ever

02-09-2007, 02:27 PM

hello! thankyou!

however, what does that mean?

guelphdad

02-09-2007, 02:37 PM

what do you think it means? What is a variable? You should know that. The error message is telling you there is an error on that line. You should have caught that before posting. As you didn't I pointed it out so change that to a variable.

Lord Emperor

02-11-2007, 10:45 PM

Newbie-friendly hint: variables in PHP must be prefixed with a dollar "$". So change

db_selected = mysql_select_db($dbname, $link);

to

$db_selected = mysql_select_db($dbname, $link);

Also, I noticed that in the end you use

mysql_close($con);

but your DB connection is $link.

neomaximus2k

02-13-2007, 09:56 AM

as Lord Emperor stated you missed out the $ on the following line
db_selected = mysql_select_db($dbname, $link);
if (!$db_selected) {

should be
$db_selected = mysql_select_db($dbname, $link);
if (!$db_selected) {