[Resolved] Displaying info from table *New to PHP*

[pipedreamer]

Hello! The answer to this may be blatantly obvious, but I've only been working with PHP for a couple of weeks now, so bear with me!

I intend to have a memberlist containing a list of members along with a link to each members profile page.

"...profile.php?userID=1"
"...profile.php?userID=2"
...etc.

When the link is clicked the userID string in the url is converted to a variable which helps me to call that users information from the 'users' table in my database.

Here is an example of my script (after connecting to db).

...............................
$result=mysql_query("SELECT * FROM users WHERE userID='$userID'", $db);

$username=mysql_result($result,"username");
$fname=mysql_result($result,"fname");
$lname=mysql_result($result,"lname");
$bio=mysql_result($result,"bio");
$avatar=mysql_result($result,"avatar");

?>
................................

When I try to print a variable, though, regardless of which variable I use, it always outputs the userID number.

What is my script missing? I hope my error is obvious to someone!

Thanks!!!!
Jamie

[cgraz]

have a look at the sticky in the Database forum entitled Database forum FAQ + Bookmarked threads Go to the second post where it says "How to retrieve and display data from database". It's just a quick introduction to retrieving info from a mysql database, so if you run into any problems post back here.

Cgraz

[pipedreamer]

Thanks cgraz. That makes perfect sense. What I'm trying to do is a little bit different though. I can't explain it much better than I already have in my first post, but I will try.

I have one profile template page (profile.php) with all of the html in place. Each user will have a profile page based upon their userID, which is set up to auto increment in the users table. For instance, I am the first user in the database, and my profile page url is "....profile.php?userID=1". The next user would have a different url based off of their userID, but the profile.php template would be the same.

I haven't found any documentation on how to populate each of these pages automatically with each users information, so I brainstormed and the best I could come up with was to use the userID in the url (which is automatically converted into the variable $userID) to call that users info from the database.

First the user is found when the $userID variable is matched to the userID field in the table. What I need help with is getting the rest of that users information from the table and outputting it to the profile page.

Does this make sense? 😛

Again, I can't seem to find any information anywhere on how to use the URL string to call information from the database. Hmm.

[cgraz]

ahhh....

Yeah, that's the same concept. Your SELECT query will need a WHERE clause (so you select only the information WHERE the id field matches that certain members id). Displaying is still the same.

So once you have your id variable in the url; something like this

www.yourdomain.com/profile.php?id=7

What you do is

$query = mysql_query("SELECT * FROM table WHERE id='" . $_GET["id"] . "'") or die(mysql_error());

$_GET is used to grab variables from the url. what is between brackets is the actual name of the variable. So if you had

profile.php?random_var=hello

To access that variable, you would use $_GET['random_var']

make sense? Let me know if you need me to expand on anything or if you run into any problems.

Cgraz

[pipedreamer]

Hello again.

Yes that really helped. I was unsure exactly how to use the $_GET command. Now I know how it works; thanks!

The problem still is this: when I try to echo/print the user information, instead of getting what I should (username, first name, etc), all it prints out is the userID number.

Here is a screen capture of part of the page. I am testing userID=16. In the places where I tried to print username, first name, last name, gender, bio, and avatar it printed out the number 16 instead.

I have a feeling this is so simple, yet I've been banging my head over this for almost two days now...

!!

[cgraz]

try changing that first part of your code to

$result = mysql_query("SELECT * FROM users WHERE userID='" . $_GET["userID"] ."'", $db); 

$row = mysql_fetch_assoc($result);

$username = $row["username"];
$fname = $row["fname"];
$lname = $row["lname"];
$bio = $row["bio"];
$avatar = $row["avatar"];

Let me know how that works for you.

Cgraz

[pipedreamer]

So tell me - did you hear my exuberant "Oh yeah!" up there in Washington?

It works (I could kiss you). Can you explain what I was doing wrong with the last script and why only the userID printed? It has been challenging keeping my head wrapped around all of this information I'm learning. But it is all very rewarding when you can get it figured out!

I appreciate your help. I'm sure there are more exciting ways for you to spend your evening, but your time really is appreciated.

Jamie

[cgraz]

no problem Jamie. What the code does is execute the query for that particular user id, fetch the results into an associative array [which is basically an array where the array keys can be words rather than numbers. For example, the column name in the db], then created the variables by using $row["column_name"] format.

Hope that makes sense.

Cgraz

[pipedreamer]

Perfect sense.

Many thanks.