also if $NAMEID is a string you select should be:
"SELECT * FROM inventory WHERE ID='$NAMEID'";

--- In [email protected], "Steve Penney" <[EMAIL PROTECTED]> wrote:
>
> Hello, I am new to this group and php mysql programming. I have
> been a MSAccess programmer for years. I have a web site hosted on 
> Yahoo where I want to query the mysql database from a form on one
> page and return query results on another. I have been trying to do
> this for two weeks and for the life of me cannot figure out what I 
am
> doing wrong. Any help would be appreciated.
> PHP Version 4.3.11
> phpMyAdmin 2.6.3-pl1
> session.auto_start = off
> 
> 
> 
> 
> 1st form - test.php
> <?php
> session_start();
> >?
> //.............
> <form method= "POST" action="GetForm.php">
> ID: <input type="text" name="first" />
> <input type="submit" />
> </form>
> //.........
> <?php
> $fname = $_POST["first"];
> $_SESSION["first"]= $fname;
> ?>
> 
> 
> 2nd form - GetForm.php
> <?php
> session_start();
> echo $_SESSION["first"];
> $NAMEID = $_SESSION["first"];
> echo $NAMEID;
> //..........
> $query = "SELECT * FROM inventory WHERE ID=$NAMEID";
> 
> ERROR MESSAGE is below.
> Query failed: You have an error in your SQL syntax; check the 
manual
> that corresponds to your MySQL server version for the right syntax 
to
> use near '' at line 1
> 
> I do not see the $_SESSION variable carrying the form input to the
> echo statement on the second form. Is this due to something not set
> properly?
> thanks,
> Steve
>






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