PHP/mySQL dynamic form help

Peteee

Hey,

Am having a bit of trouble generating a dynamic form in php, getting info from a mySQL database

The form isnt displaying anything (i.e. its blank) although it does 'go down' the correct number of spaces. Hope you know what i'm on about here!

// Select fields in the bugid table to be displayed
		$bugidQuery = "SELECT BugID FROM BugFind";
		$bugidResult = mysql_query($bugidQuery);
		
	
		// Determine the number of Bugs
		$bugidNumber = mysql_numrows($bugidResult);

		// Print the Bug names as a drop-down menu

		print "Pick the bug that you wish to modify : ";

		print "<select name=\"bug\">
    	<option value=\"\">Select an Bug</option>";

		for ($i=0; $i<$bugidNumber; $i++) {
        	$bugidName = mysql_result($bugidResult,$i,"bug");
       	 
        print "<option>$bugidName</option>";
}
		print "</select><p>";
		

The strange thing is i have nearly identical code below, and it works fine!!!

i.e. instead of blanks, it displays stuff

//get list of features leave it up to the user to pick correct ones
	//only makes sense to pick active features when logging a bug

	$featureQuery = "SELECT feature FROM Features where status='Active'";
	$featureResult = mysql_query($featureQuery);

	//get no. of features
	$featureNumber = mysql_numrows($featureResult);

	//make drop down menu
	print "Select the Feature: ";

	print "<select name=\"feature\">
    <option value=\"\">Select a Feature</option>";

	for ($i=0; $i<$featureNumber; $i++) {
 	 $featureName = mysql_result($featureResult,$i,"feature");
  
  print "<option>$featureName</option>";
}
print "</select><p>";
		

To me they look identical, but then it is driving me crazy so.....

It could also be a probelm with the database???

Thanks for any help in advance, any more info needed and i'll gladly show you

kenmc

you're pulling X elements from the database correectly, but you're not accessing them correctly in the results..
check this....
$bugidQuery = "SELECT BugID FROM BugFind";
this will result in a small table looking like this:
BugID

---

123
1231
3321
213
...

$bugidName = mysql_result($bugidResult,$i,"bug");
but here you're looking for things called "bug" in the results, but there are none - need to change "bug" here to "BugID".
K

daymobrew

While I don't know much SQL, I know some debugging techniques.

Have you run the problem query from a console to ensure it returns data? (I'm guessing yes).
Add return code checks to ensure mysql functions are not having problems.
Maybe "print mysql_error()" every few lines.

I was reading:
http://www.php.net/manual/en/function.mysql-result.php

[Aside: You should ensure $bugidResult is > 0 before printing 'select' tags]

kenmc

daymo,
the fact that "it does 'go down' the correct number of spaces" indicates that s/he's getting data back, but is just not pulling it from the result array correctly (asking for apples when there are only oranges there....)

Peteee

Thanks very much, i knew it would be something trivial like that!

My eternal gratitude, hopefully will not make a stupid mistake like that again!

kaimera

ah pete, you should have asked me! I haven't had time to look at it properly yet

kenmc

u're welcome. 2nd pair of eyes often spots it....

Peteee

Yep, unfortunatly 3 pairs of eyes didnt spot it our house

Ah well 2 heads are better then 3!