Success with EasyPHP!

I have been very successful with EasyPHP and Win.2000 but I'm still having problems. The first problem I corrected by switching text editors. I started with Notepad but it kept sticking a .txt extension on the end and that drove me crazy for a few days. Once I switched and started writing real .php docs I was in-like-Flin. Thanks for everyones help.

Now I am having trouble connecting to mySQLdatabase, although I have no trouble using PHPmyadmin to create tables, databases, etc. I get a parse error with the following code:

first what I'm using... apache 1.3.24.+ , PHP 4.2.0 and mysql (latest)...

Here is the code:

<? PHP
$db = @mysql_connect("localhost", "root", "");
mysql_select_db("myfirstdb",$db);
if ($submit) {
// here if no ID then adding else we're editing
if ($id) {
$sql = "UPDATE booksbd SET booktitle ='$booktitle',author='$author',publisher='$publish er',summery ='$summery',inputby ='$inputby', WHERE id=$id";
} else {
$sql = "INSERT INTO booksbd (booktitle,author,publisher,summery,inputby) VALUES ('$booktitle','$author','$publisher','$summery', '$inputby')";
}

* I wanted to enclose the above code in red, php code, but I don't know how...


I'm so close, can anyone help?
Thanks.

whoa! a few big problems, here.

Firstly im guessing that you used to use 4.0 and have only made the switch reciently to 4.2.

Ok firstly, if your connecting to your database server via phpmyadmin without any hassels, then everything is set up correctly.

Im asuming that you have the right connection details, so lets move on

if ($submit) {
// here if no ID then adding else we're editing
if ($id) {
$sql = "UPDATE booksbd SET booktitle ='$booktitle',author='$author',publisher='$publish
er',summery ='$summery',inputby ='$inputby', WHERE id=$id";
} else {
$sql = "INSERT INTO booksbd (booktitle,author,publisher,summery,inputby) VALUES ('$booktitle','$author','$publisher','$summery', '$inputby')";
}

here lies your problem, firstly you want to insert or update data into the database, currently all your doing is setting a value of a $VARIABLE. you actually need to query the database like this



next im asuming that this page is loaded after you hit the submit button, in which case, it wont get past the

if ($submit) {

line.

4.2.x uses superarray globals, so basically this means, that you cant call a post $variable as $submit.

you will need to do it like this



next problem

your queries are calling $var, from POST variables

you will also need to use $_POST

but you will need to place them in curly brackets {} since your setting them in a variable

an example

$sql = "SELECT * FROM books WHERE field = {$_POST['id']}";

something like that

Thanks, I'll get to the changes. I'll post another reply once these are in place and I test it again! Thanks again.