Stupid newbie question

[Renegade]

Hi everyone, I am new to PHP, but not to programming. I am well versed in C, so the learning process is going quite smoothly and quickly. More so than I would have thought

But I digress....

In plain simple english, here is the query I am trying to perform on the MySQL server:

"What is the id of the person whose first name is ____"?

Here is my code:

$sql = "SELECT id FROM users WHERE firstname='$firstname';";
$result = mysql_query($sql);
$userrow = mysql_fetch_array($result);
$userid = $userrow["id"];

The $firstname var gets its value from the user who selects the name of the person. I have verified that $firstname contains the desired data.

It seems that $userid is always 0, no matter what.

What am I doing wrong here? Suggestions?
Thanks

[exbabylon]

Try this:

$sql = "SELECT id FROM users WHERE firstname='$firstname'";
$result = mysql_query($sql);
while($row = mysql_fetch_array($result)){
      echo $row['id'];
} 

[exbabylon]

[pedro_gb]

Are you sure id is a column on the database? That has already happened to me.

[Karl]

The problem is that you have a ; at the end of the SQL query inside the ". You don't need it as PHP automatically adds it in for you. So the end of your query should look like '"; not ';";

[Renegade]

wow, thanks for the quick replies...

Yes, id does exist as a column in the DB....lol

I copied your code, exbabylon, and pasted it into mine, still no workie. In fact, it didn't even echo anything out to the screen. The screen was blank as if nothing was echoed at all

Removing semi colons didn't help either.

Anything else I might try?

[petesmc]

Try:

<?php
mysql_connect("host", "username", "pass");
mysql_select_db("databasename");
print $firstname; # Check whether it really exists!
$sql = "SELECT id FROM users WHERE firstname='$firstname'";
$result = mysql_query($sql);
$userrow = mysql_fetch_array($result);
$userid = $userrow["id"];
print $userid; #Try and print it
?>

[freddydoesphp]

Pete, I hate to sound like a nag, but I don't want anyone to get confused when they read your code

$print $firstname;

should actually be

print $firstname;

[petesmc]

Opps....DIdn't notice that...THanks...

[Renegade]

ok, figured it out.

I was getting input from user from a selection list containing first and last names, seperated by a comma. I used explode() to extract each seperate name....whitespace and all

I used trim() on $firstname and voila! worked perfectly.

DOH!

thanks everyone....