[NeverMind]

ok .. I just wrote my first ever php script .. but it's showing a parser error in the IF statment ..

Parse error: parse error, unexpected T_IF in
c:\program files\apache group\apache\htdocs\scripts\addition.php on line 31

and here is my php code :

PHP Code:

<?php

/*first get all feilds values in here ..*/

    $title=$HTTP_POST_VARS['title'];

    $bookid=$HTTP_POST_VARS['bookID'];

    $authf=$HTTP_POST_VARS['authf'];

    $authl=$HTTP_POST_VARS['authl'];

    $price=$HTTP_POST_VARS['price'];



/*now we Clean in the entries from spaces , to make

 sure the DB don't get screwed up ..*/

    $title=trim($title);

    $bookid=trim($bookid);

    $authf=trim($authf);

    $authl=trim($authl);

    $price=trim($price);



/*another cleaning process ,, but this time to make sure 

no special chars would hurt the DB */

    $title=addslashes($title);

    $bookid=addslashes($bookid);

    $authf=addslashes($authf);

    $authl=addslashes($authl);

    $price=addslashes($price);



/*now we should specify the MySQL connection info ..*/

    $db_name="bookdb";

    $server="localhost";

    $db_user="saleh";

    $db_pass="hidden"



/*check if user completed the form or not ..*/

if (!isset($title) || !isset($bookid) || !isset($authf) || !isset($authl) || !isset($price))

                           {

echo "Sorry but you did not complete all fields .. please go back and fill ALL fields";

                        }

/*now we will try to connect to MySQL database .*/

                            else

        {  

            echo "now we are trying to connect to MySQL ..";

        }

    mysql_connect($server , $db_user , $db_pass)

    or die("Could not connect: " . mysql_error());

    print ("Connected successfully");



    mysql_db_select($db_name);

    mysql_query("INSERT INTO books (title, bookID, authF, authL, price)

    VALUES ($title, $bookid, $authf, $authl, $price)");



        $result=mysql_query("SELECT * FROM books", $db);

        $record_total=mysql_num_rows($result);

        echo "Total of books Record in the database is :" .$record_total .;

?>

the varibles values should be passed via form submittion .. the form is plain HTML ....
you know I love php but after this thing I started to hate it

[Godsmack]

You forgot the semicolon on the line above the IF statement.

PHP Code:

$db_pass="hidden" 

Randy

[NeverMind]

ok this is done ..
also mysql_select_db(); is done .. now it's showing this :

now we are trying to connect to MySQL ..Connected successfully
Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in c:\program files\apache group\apache\htdocs\scripts\addition1.php on line 46

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in c:\program files\apache group\apache\htdocs\scripts\addition1.php on line 47
Total of books Record in the database is :

[NeverMind]

ok ... this is done ..
thanks the legend [Unknown] (he is a friend of mine poped up in MSN messanger after I posted this topic)
just for the record here are the problems in my code :

PHP Code:

$db_pass="hidden" 

should have a semicolon

PHP Code:

mysql_query("INSERT INTO books (title, bookID, authF, authL, price)   VALUES ($title, $bookid, $authf, $authl, $price)"); 

varibles should be in quote ' '

PHP Code:

$result=mysql_query("SELECT * FROM books", $db); 

I put an undefined varible ($db) which isn't needed anyway ...
(I don't know how I did this stupid mistake :-[)

PHP Code:

echo "Total of books Record in the database is :" .$record_total .; 

the . (dot) before ; is wrong ..

that's all .. now it's working perfectly ..
thanks again for [Unknown]

[Godsmack]

change line 46 to:

PHP Code:

mysql_select_db($db_name); 

instead of mysql_db_select

Randy