[RESOLVED] PHP Dropdown Menu Help

OneCrispy

Hi I'm New to PHP and I need some help with this code which I got off a PHP tutorial of how to create a PHP Drop-Down Menu using Javascript and PHP.

However I seem to be reciving some errors, I have followed the code exactly and also written some comments on the tutorial website but it was not much help.

Here is the code for both the update.php and sucess.php I hope you will be able to help. Thanks.

Update.php:

<?php

require "connection.php";
echo "

function fillCategory(){ 

";
$q1=mysql_query("select * from `group`");
echo mysql_error();
while($nt1=mysql_fetch_array($q1)){
echo "addOption(document.drop_list.Main, '$nt1[group]', '$nt1[group]');";
}
function SelectCat(){

removeAllOptions(document.drop_list.Category);
addOption(document.drop_list.Category, "", "Category", "");

$q2=mysql_query("select distinct(`group`) from category");

while($nt2=mysql_fetch_array($q2)){

echo "if(document.drop_list.Main.value == '$nt2[group]'){";
$q3=mysql_query("select category from category where `group`='$nt2[group]'");
while($nt3=mysql_fetch_array($q3)){
echo "addOption(document.drop_list.Category,'$nt3[category]', '$nt3[category]');";

}
echo "}";
}
function SelectSubCat(){

removeAllOptions(document.drop_list.SubCat);
addOption(document.drop_list.SubCat, "", "SubCat", "");


$q4=mysql_query("select distinct(`category`) from subcategory");

while($nt4=mysql_fetch_array($q4)){

echo "if(document.drop_list.Category.value == '$nt4[category]'){";
$q5=mysql_query("select subcategory from subcategory where `category`='$nt4[category]'");
while($nt5=mysql_fetch_array($q5)){
echo "addOption(document.drop_list.SubCat,'$nt5[subcategory]', '$nt5[subcategory]');";

}
echo "}"

?>

Sucess.php:

<?php
function removeAllOptions(selectbox)
{
var i;
for(i=selectbox.options.length-1;i>=0;i--)
{
selectbox.remove(i);
}
}

function addOption(selectbox, value, text )
{
var optn = document.createElement("OPTION");
optn.text = text;
optn.value = value;

selectbox.options.add(optn);
}
[

?>

Tutorial Website: http://teamtutorials.com/web-development-tutorials/pulling-drop-down-values-from-a-database-using-javascript-and-php

paulnaj

Hi, what errors are you receiving - it's bit much to have to back engineer the code!!

P. 😉

OneCrispy

Hi Paulnaj,

I seem to be reciving a Parse error: syntax error, unexpected $end and a Parse error: syntax error, unexpected ')', expecting '&' or T_VARIABLE

Thanks for your help

dagon

that's really weird code, why are you trying to echo out the php?

OneCrispy

I dont know. Like I say I got it off a tutorial. I'm not sure what you mean

dagon

wheres the tutorial ?

OneCrispy

Tutorial Website: http://teamtutorials.com/web-development-tutorials/pulling-drop-down-values-from-a-database-using-javascript-and-php

paulnaj

Actually I think it's pumping out javascript, no?

OneCrispy

Yes it is on a seperate page,

Code:

<html>
<head>
<title>Javascript Form Updater</title>
<script language="javascript" src="update.php"></script>
</head>

<SELECT NAME="Main" onChange="SelectCat();" >
<Option value="">Main</option>
</SELECT> 
<SELECT id="Category" NAME="Category" onChange="SelectSubCat();">
<Option value="">Category</option>
</SELECT>
<SELECT id="SubCat" NAME="SubCat">
<Option value="">SubCat</option>
</SELECT>
</form>

</body>

</html>

Dont think there is anything wrong with that code.
Thanks

dagon

there are a number of lines from the tutorial that you have missed in your code.

OneCrispy

Which Lines?

dagon

check it, i found a couple in 1 minute of looking

OneCrispy

Nope, no difference still reciving same errors,

Thanks for taking the time to look though

dagon

then post your new Update.php because it cant be the same as the one above.

OneCrispy

<?php
require "connection.php";
echo "

function fillCategory(){

";
$q1=mysql_query("select * from group");
echo mysql_error();
while($nt1=mysql_fetch_array($q1)){
echo "addOption(document.drop_list.Main, '$nt1[group]', '$nt1[group]');";
}
function SelectCat(){

removeAllOptions(document.drop_list.Category);
addOption(document.drop_list.Category, "", "Category", "");

$q2=mysql_query("select distinct(group) from category");

while($nt2=mysql_fetch_array($q2)){

echo "if(document.drop_list.Main.value == '$nt2[group]'){";
$q3=mysql_query("select category from category where group='$nt2[group]'");
while($nt3=mysql_fetch_array($q3)){
echo "addOption(document.drop_list.Category,'$nt3[category]', '$nt3[category]');";

}
echo "}";
}
function SelectSubCat(){

removeAllOptions(document.drop_list.SubCat);
addOption(document.drop_list.SubCat, "", "SubCat", "");

$q4=mysql_query("select distinct(category) from subcategory");

while($nt4=mysql_fetch_array($q4)){

echo "if(document.drop_list.Category.value == '$nt4[category]'){";
$q5=mysql_query("select subcategory from subcategory where category='$nt4[category]'");
while($nt5=mysql_fetch_array($q5)){
echo "addOption(document.drop_list.SubCat,'$nt5[subcategory]', '$nt5[subcategory]');";

}
echo "}";
}

dagon

the tutorail


echo "addOption(document.drop_list.Main, '$nt1[group]', '$nt1[group]');";
}
?>
}
function SelectCat(){

your code

echo "addOption(document.drop_list.Main, '$nt1[group]', '$nt1[group]');";
}
function SelectCat(){

OneCrispy

Still arises with same error: Parse error: syntax error, unexpected $end in /home/amateurg/public_html/update2.php on line 48

dagon

post your current version

OneCrispy

<?php
require "connection.php";
echo "

function fillCategory(){

";
$q1=mysql_query("select * from group");
echo mysql_error();
while($nt1=mysql_fetch_array($q1)){
echo "addOption(document.drop_list.Main, '$nt1[group]', '$nt1[group]');";
}
function SelectCat(){

removeAllOptions(document.drop_list.Category);
addOption(document.drop_list.Category, "", "Category", "");

$q2=mysql_query("select distinct(group) from category");

while($nt2=mysql_fetch_array($q2)){

}
echo "}";
}
function SelectSubCat(){

removeAllOptions(document.drop_list.SubCat);
addOption(document.drop_list.SubCat, "", "SubCat", "");

$q4=mysql_query("select distinct(category) from subcategory");

while($nt4=mysql_fetch_array($q4)){

echo "if(document.drop_list.Category.value == '$nt4[category]'){";
$q5=mysql_query("select subcategory from subcategory where category='$nt4[category]'");
while($nt5=mysql_fetch_array($q5)){
echo "addOption(document.drop_list.Main, '$nt1[group]', '$nt1[group]');";
}
?>
}
function SelectCat(){

dagon

so you make no changes, ignore most of the tutorial, ignore the advise given here and expect it to work, well good luck to ya.