[C_coder]

hey guys...

I'm having a problem with the <option> and <select>...I go to the table to fetch the data but I can't seem to put it on the "combobox", combobox comes empty!

Any idea what it might be?

here is the code I'm using:

echo "<select>";
		
                
                
                echo "<option"; if($row['emplacement'] == 1) echo "value ='1'>place 1</option>";
                echo "<option"; if($row['emplacement'] == 2) echo "value ='2'>place 2</option>";
                echo "<option"; if($row['emplacement'] == 3) echo "value ='3'>place 3</option>";
                
                echo "<option"; if($row['emplacement'] == 4) echo "value ='4'>place 4</option>";
                echo "<option"; if($row['emplacement'] == 5) echo "value ='5'>place 5</option>";
                echo "<option"; if($row['emplacement'] == 6) echo "value ='6'>place 6</option>";
		echo "</selected>";

[RudiVisser]

Well yes, did you look at the produced HTML?

It will be something like this:

<select><option<optionvalue='2'>place 2</option><option<option<option</selected>

Does that look right to you?

It should be (provided I understand what you're doing):

<select>
    <option value="1">place 1</option>
    <option value="2" selected="selected">place 2</option>
    <option value="3">place 3</option>
</select>

1; Add a space
2; Change <selected> to <select> because that is the tag you're creating
3; Always output the value because at the moment you have a lot of malformed HTML being output
3; I think that you could be missing the point of what this is doing, you're going to have a lot of options without values if you use this code. Your code, I think, should look more like this:

?><select>
    <option value="1"<?php if ($row['emplacement'] == 1) { ?> selected="selected"<?php } ?>>place 1</option>
    <option value="2"<?php if ($row['emplacement'] == 2) { ?> selected="selected"<?php } ?>>place 2</option>
    <option value="3"<?php if ($row['emplacement'] == 3) { ?> selected="selected"<?php } ?>>place 3</option>
    <option value="4"<?php if ($row['emplacement'] == 4) { ?> selected="selected"<?php } ?>>place 4</option>
    <option value="5"<?php if ($row['emplacement'] == 5) { ?> selected="selected"<?php } ?>>place 5</option>
    <option value="6"<?php if ($row['emplacement'] == 6) { ?> selected="selected"<?php } ?>>place 6</option>
</select><?php

or even better (short tags/PHP 5.4 permitting):

?><select>
    <?php for($i = 1; $i < 7; $i++): ?>
        <option value="<?= $i ?>"<?= ($row['emplacement'] == $i) ? 'selected="selected"' : '') ?>>
            place <?= $i ?>
        </option>
    <?php endfor ?>
</select>

This post has been edited by RudiVisser: 07 March 2012 - 01:10 AM

[C_coder]

You are right...My code was a bit "confussing" for the least...thanks for the explenation it really helped me understand the code...I had it all wrong in my head!

[C_coder]

Not I'm having another problem with this combobox.

when I click in the submit button it goes to my page "update_ligne.php" but when I declare a variable to have the value from emplacement it gives me this
¨

( ! ) Notice: Undefined index: emplacement in C:\wamp\www\PHP\update_ligne.php on line 32
Call Stack
#	Time	Memory	Function	Location
1	0.0375	374496	{main}( )	..\update_ligne.php:0

and this is my ligne 32

$emplacement = 	($_POST['emplacement']);

Can't seem to find error cus I I'm using pretty much the same code as in my creation form and there it is working fine!!

Any ideas?

[Macjohn]

Try putting it like this:

$emplacement =  (isset($_POST['emplacement']));

[C_coder]

Thanks, it solved my problem with ligne 32!.

But now I'm having another error. I think it is my UPDATE function!

[Macjohn]

show us your UPDATE code and we may be able to help you!

[C_coder]

$qry = "UPDATE INTO computers(machine, ip, emplacement) VALUES('$machine','$ip','$emplacement')";
	$result = @mysql_query($qry);
	
	//Check whether the query was successful or not
	if($result) {
		header("location: index.php");
	
		exit();
	}else {
	echo('Query failed');
		die("Query result failed");
	}

[C_coder]

i tried this too...

$qry = "UPDATE computers SET machine='$machine' ip='$ip'";
	$result = @mysql_query($qry);

Not working

[RudiVisser]

mysql_error()

Not to be funny, but I think you should go learn PHP basics.

[C_coder]

Sorry but I don't understand what you mean...
I know I still need to learn alot but where should i put that mysql_error()

top, bottom

its ok...found it...

[JackOfAllTrades]

RudiVisser, on 07 March 2012 - 05:22 AM, said:

mysql_error()

Not to be funny, but I think you should go learn PHP basics.

Totally agree. Seems like 90% of the people posting here have absolutely NO concept of the basics at all, and NO logical or debugging skills of their own. It's extremely frustrating.

Like this:

Quote

I know I still need to learn alot but where should i put that mysql_error()

Seriously??? No sense of logic at ALL!

"Here's a function that will print out an error."

"Hurr durr...where do I put it?"

Where the HELL do you THINK you'd put a function that prints an ERROR???

Sorry if you think that's harsh, but that's just so ridiculous on its face. I'm guessing you're just copying and pasting your way to a solution.

[Macjohn]

Ok, I see you still didn't found your error (cus if you did try to post it here to help some other people in the future!)

try to write it like this:

$qry = "UPDATE computers SET machine='$machine', ip='$ip'";

You forgot a small thing ","

cya mate

[Macjohn]

oh, and try to read before asking...google is a really good friend!
Search first, ask later!

But don't take it the wrong way!

Cya

[C_coder]

Ok, sorry for before, I'll try to search a bit better next time I have a stupid question!

Oh, and I found the answer by myself but thanks again for the time!