Clinton

Are you posting a form at all or just trying to query the location if the page switches?

I've got a table that lists office locations by city and state. Sometimes there are several offices in one city. Right now I am using the following code to pull up the citys City: <select name="city" onchange='this.form.submit()'> <?php $sql = "SELECT city FROM thelist WHERE state = '$state' ORDER BY city"; $rs = mysql_query($sql); while($row = mysql_fetch_array($rs)) { extract($row); echo "<option value='$city'>$city</option>"; } ?> </select> But if there is three offices in one city then it will echo out that same city 3 times in the drop down list. What's the best way to correct this so it only shows once?

Thanks Cy, that worked perfectly.

Thanks Gevans, that worked. Now one more question. If the state variable hasn't been selected I want them to be able to view all states. Do I have to do an if isset statement or can I use a * or a wildcard variable of somesort? I've tried both * and % to no avail.

I didn't think it was that complicated but everytime I add another variable it seems to not work. This time it's the state variable. I added it and my query is pretending like it's not there and displaying the rest of the result. ANy ideas? $sql = "SELECT * FROM $tbl_name WHERE dtype = '$dtype' OR dtypewc = '$dtype' AND dmajorp = '$smajor' OR dmajorc1 = '$smajor' OR dmajorc2 = '$smajor' OR dmajorc3 = '$smajor' AND state = '$state' ORDER BY jpted LIMIT $start, $limit";

So if bob logs in then I store a 'yes', for example, in a table specifically designed to indicated bob's logged in status. But again, what happens if he closes his browser without logging out? How does the db update then? I thought I was just setting sessions but my sessions weren't dying when I unset them so I used the whole cookie deal and wah-lah it works. See http://www.phpfreaks.com/forums/index.php/topic,232287.0.html

Well regardless of cookie time if a browser is closed the sessions generally kill, at least most websites I have been on. Mine's not doing that and I prefer it that way. Other than that they could stay logged in for eternity, which is fine with me hence the unusually long time.

Ok, thanks guys, yea, both of those work. :-)

I did the code below because I'm going to replace dtype = * with dtype = $variable and I wanted to get it to work... which it's not. Basically, I want to be able to narrow down the search by dtype AND the dmajors but if there is no specific dtype selected I want to show all of the dtypes. Make sense? <?php $query = "SELECT COUNT(*) as num FROM $tbl_name WHERE dtype = * AND dmajorp = '$smajor' OR dmajorc1 = '$smajor' OR dmajorc2 = '$smajor' OR dmajorc3 = '$smajor' ORDER BY jpted"; ?>

Ok, when I click the logout button this gets executed and the session ends. if (isset($_COOKIE[session_name()])) { setcookie(session_name(), '', time()-42000, '/'); However, if the browser closes, or even when the computer restarts, the session still remains if I do not logout. How do I fix this? Do I have to put the setcookie on every page...???

Apparently all sorts of ways I could have done it. LoL. Thanks Premiso... once again. :-)

It's an option on a form, text type, then the form INSERTS all info into the db. It's actually being populated by SELECT then extract_row().

Ok, Premiso... I did not enter anything into that line (dtypewc) so why did it "set"? How does something "set"? I looked up in the manual but it doesn't say how a variable set's. Thanks for the empty link btw.

I have this... <?php if (isset($dtypewc)) { ?> <tr><td id='header'> Will Consider a <?php echo $dtypewc; ?>. </td></tr> <?php } ?> Now, my $dtypewc variable is empty but it's still returning the "Will Consider a"... Any ideas why? Thanks.

I've been scouring the internet up and down and have found a lot of pieces about mysql, php, and pictures and have tried to put them together to form my code.... but... it's failing. So a couple of questions... I've got this on the page (we will call showimage.php) I want to show the picture: <h3><img src="getimage.php?name=<?php echo $username ?>" width=100 Height=125></h3> I've got this on the getimage.php page: <?php include 'connect/project.htm'; $name = $_GET['name']; $result=mysql_query("SELECT clogo FROM employers WHERE username='$name'") or die(mysql_error()); $row=mysql_fetch_object($result); Header( "Content-type: image/png"); echo $row->content; ?> Now, the showimage.php page is just displaying a broken link. The getimage.php page is displaying the text "http://localhost/project/getimage.php?name=hunna03" as an image and not the actual image. So the questions... 1) I am saving images as a BLOB in mysql. How do I know they are saving correctly? I can't find any way to see them in MySQL. 2) What am I doing wrong with this code? Thanks.

Can you show us your code?

You're right, my apologies.

How long is this supposed to go on?

Got it. Just had by variable names wrong. :-p Thank you.

No, gevans, you're hiding the button. He wants: <form> <input type="hidden" name="uid" value="<?php echo $rows['UID']; ?>" /> <input type="submit" name="edit" value="EDIT"> Then on the next form you justget $_POST['uid'] and go from there.

I was messing with that but it's not working... let me play with it a little bit more.

What exactly is the question here? What are you trying to do?

Let's say I've got three fields that keep track of a persons skills. These are Bobs skills... Field 1 is his primary skill, say a Carpenter. Field 2 is his secondary skill, say a Hair Dresser. Field 3 is his tertiary skill, say a Window Washer. These are Joes skills... Field 1 is his primary skill, say a Hair Dresser. Field 2 is his secondary skill, say a Carpenter. Field 3 is his tertiary skill, say a Window Washer. Now lets say I wanted to pull up all the people who are qualified to be a carpenter. How do I do this? Right now I've got: SELECT * FROM thelist WHERE pskill = 'carpentry' But this only lists the ones who have the primary skill of carpentry. If I do: SELECT * FROM thelist WHERE pskill = 'carpentry' AND sskill = 'carpentry' then it leaves out anybody who just has the primary skill (or secondary if I include the third). How do I make it so it finds all carpenters in any of those three fields without excluding everyone if one doesn't have it?

If you know exactly what category then yes, that's correct.

Well, I wouldn't say that I've got it but I would say that it makes more sense. :-) Thanks for taking the time to explain it.