selected drop-down option

[[deleted]]

I'm trying to create some code that will display the value in my MySQL database as the selected value in a drop-down menu. All of the other options should be displayed as well.

Here is what I have. The code below displays all of the options, but not what is selected.

Some assistance would be appreciated.

Thanks.

<?php 
		$qGetSubCategoryName = "SELECT SubCatID,name 
		FROM tblSubCategory";
		$rGetSubCategoryName = mysql_query($qGetSubCategoryName); 

		echo "<select name=\"newSubCatName\">"; 

		while ($GetSubCategory_row = mysql_fetch_array($rGetSubCategoryName)) { 

		$SubCatID = $GetSubCategory_row['SubCatID']; 
		$SubCatName = $GetSubCategory_row['name'];
		$SubCatOption = "<option value=\"".$SubCatID."\""; 

		if ($GetSubCategory_row == $SubCatCatID) { $SubCatOption = $SubCatOption." selected>$SubCatName</option>"; 
		} else { 
		$SubCatOption = $SubCatOption.">$SubCatName</option>"; 
		} echo "$SubCatOption"; 
		} echo "</select>"; 


		?>

[xblue]

what is that $SubCatCatID you use in your comparison?
I only see $SubCatID elsewhere in the code.

Typing error?

[[deleted]]

Yes, it's a typo. It should be $SubCatID. I changed it and still no luck.

[vfwood]

hi

it looks like a simple typo to me (i haven't ran your code but looked through it).

the line

if ($GetSubCategory_row == $SubCatCatID) 

refers to

$SubCatCatID

surely this should be

$SubCatID

hope this helps let me know if it doesnt.

regards

vince.

p.s. i 'd use another name for the 'name' field you refer to - it may not be a mysql keyword but does not say anything about what it is the name of.

[vfwood]

sorry - didnt see other posts before i submitted for some reason.

doh! (shoot me!)

[xblue]

I don't not really get which value you would like to have selected.
however, with ($GetSubCategory_row == $SubCatID) you are comparing an array to one of its elements which doesn't seem to make much sense to me.

[[deleted]]

The value I'm trying to select is SubCatID. I'm trying to get it look at the value of tblPic.fk_SubCatID and say that if fk_SubCatID has a corresponding value in tblSubCategory then set the value of tblSubCategory.SubCatID as selected in the drop-down menu.

Not that I have thought about this a little more I realize that that is not what the code says. I guess I should RTFC (Read The Fine Code).

I could still use some help modifying what I have to what I want it to be.

I hope this clears up the confusion.

[keith73]

I need to do this a lot and I've found that creating a function to create the select menu works well. I often need to call it in more than one place.

Here's an example of a function that I would use. Note that the parameter being passed, $selected is set to 0, that is the default if no value is passed to the function. If you are looking to have a particular option selected, pass the id to the function.

function CategoryDDmenu($selected=0)
{
$Query = "SELECT id,name FROM table_name ORDER BY sort_key ASC";
$Result = mysql_query($Query);
if($Result){
	if(mysql_num_rows($Result) <= 0){ print "NO RESULTS FOUND!"; }
	else {
		print "<select name=\"fieldname\">\n";
		if($selected == 0){
			print "<option value=\"0\" selected> -- SELECT AN OPTION -- </option>\n";
		}
		while($RowData = mysql_fetch_array($Result)){
			print "<option value=\"" . $RowData['id'] . "\"";
			if($selected == $RowData['id']){ print " selected"; }
			print ">" . $RowData['name'] . "</option>\n";
		}//END while
		print "</select>\n";
	}
} else {
	print "DATABASE QUERY ERROR!";
}//END if
}//END function CategoryDDmenu

Hope this helps,

• Keith

[[deleted]]

Thanks!

How can I modify your code so that it takes into account that I'm looking at two different tables for values? The options for the drop-down list come from tblCategory, but the selected value is the foreign key value in tblPic.

[keith73]

The function takes a parameter called selected. It can be set to string or number, and then you just need to modify the if statement that checks for the selected value. Pass the foreign key in as the selected parameter.

The other option is to modify the query to select values from 2 tables.

• Keith