[[SOLVED] css within if statement]

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

[[SOLVED] Best Practice ?]

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?

[[SOLVED] Complicated Select query]

Thanks Cy, that worked perfectly.

[[SOLVED] Complicated Select query]

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.

[[SOLVED] Complicated Select query]

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";

[setcookie ()]

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

[setcookie ()]

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.

[[SOLVED] WHERE ?]

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

[[SOLVED] WHERE ?]

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"; ?>

[setcookie ()]

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...???

[isset help]

Apparently all sorts of ways I could have done it. LoL.

 

Thanks Premiso... once again. :-)

[isset help]

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().

[isset help]

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.

[isset help]

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.

[Picture Problems]

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.

[passing value with button....pls help]

Can you show us your code?

[passing value with button....pls help]

You're right, my apologies.

[Loop of tuesdays and saturdays]

How long is this supposed to go on?

[[SOLVED] WHERE ?]

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

[passing value with button....pls help]

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.

[[SOLVED] WHERE ?]

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

[passing value with button....pls help]

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

[[SOLVED] WHERE ?]

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?

[php id?]

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

[[SOLVED] SELECT COUNT ?]

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.