passing variables from a form and displaying them

timmybuck

Hi guys, i have a little problem. I have a drop down menu and i would like it so that when someone chooses a category from my drop down menu that on the next page the events are displayed for that category, so the might choose sport as the category and the following page would display sports events.

At the momment i have an category table with categoryid and category.

i have an events table with a list of all the events.

On the drop down menu page. This is my code

print("<select name=\"category\" size=1>\n");

print("<option value=\"b\">Select");
print("<option value=\"b\">-----");
while($row = mysql_fetch_array ($event_categories))
{
print("<option value=\"$row[category]\">$row[category] \n");
}
print("</select>\n");

how should the code look on my next page to display the events table

CaptianChaos

First, you probably want to do this:

while($row = mysql_fetch_array ($event_categories))
{
print("<option value=\"{$row['id']}\">{$row['category']} \n");
}

Then, on the next page:

$sql = "SELECT * FROM whateverTable WHERE categoryID='{$_POST['category']}'";
$result = mysql_query($sql);
while ($row = mysql_fetch_array($result)) {
print_r($row);
}

timmybuck

Hi ok i made changes u said

print("<select name=\"category\" size=1>\n");

then on the next page i typed this

mysql_connect("localhost","phpuser","phpuser03") or die ("Database connection error");

mysql_select_db('booking')

$sql = "SELECT * FROM event WHERE categoryID='{$_POST['category']}'";
$result = mysql_query($sql);
while ($row = mysql_fetch_array($result)) {
print_r($row);
}

but i get an error message saying there is an error on this line
$sql = "SELECT * FROM event WHERE categoryID='{$_POST['category']}'";

can u see the error the error message is

Parse error: parse error, unexpected T_VARIABLE

shadypalm88

try

$sql = "SELECT * FROM `event` WHERE categoryID='".$_POST['category']."'";

timmybuck

damn still get the same response

Parse error: parse error, unexpected T_VARIABLE in

hmmm i wonder what im doing wrong.

The categoryID field is in the event table, so i dont know why it isn't working

shadypalm88

I need to see the rest of the error to see what's wrong (unexpected T_VARIABLE in what?).

Oh, and post the current file with all the changes, that'll help too.

timmybuck

ok thanks for your help.

This is my drop down menu page

<body>

<?php

mysql_connect('localhost', 'phpuser', 'phpuser03')or die ("Database connection error");

mysql_select_db('booking');

//gets all the available events from the events table.

$event_categories=mysql_query("SELECT * FROM event_categories ")

OR DIE("Unable to find events. Error: ".mysql_error());

print("<select name=\"category\" size=1>\n");

</form>
</body>

</html>

this is the page im tryoing to get results displayed in

<html>

<body>

<?php
$categoryID=$_GET['categoryID'];
$categoryID = addslashes('$categoryID');

mysql_connect("localhost","phpuser","phpuser03") or die ("Database connection error");

mysql_select_db('booking')

$sql = "SELECT * FROM event WHERE categoryID='".$_GET['category']."'";
$result = mysql_query($sql);
while ($row = mysql_fetch_array($result)) {
print_r($row);
}

?>
</body>
</html>

shadypalm88

I see two problems. One is causing your parse error: you forgot a semicolon (😉.

mysql_select_db('booking') should be mysql_select_db('booking');

Also, there's a problem with $categoryID = addslashes('$categoryID');

There's a difference between using single-quotes (') and double-quotes (") in PHP. Double quotes will replace any variables (like $categoryID) in them with their values. Single quotes won't. What that line actually does is add slashes to the word "$categoryID", not the actual categoryID.

Plus, when you're just using a variable, you don't need quotes at all. That line should read:
$categoryID = addslashes($categoryID);

timmybuck

thanks for the explanation

my code looks like this now

$categoryID=$_GET[categoryID];
$categoryID = addslashes($categoryID);

mysql_connect("localhost","phpuser","phpuser03") or die ("Database connection error");

mysql_select_db('booking');

but now i get no error rmessage but a blank page

we're getting closer

shadypalm88

OK, here's the only problem I can see that would cause that. On your first page, you're (sort of) sending the information as POST, but trying to retrieve then on the second page using GET. I say sort of because you spelled method wrong.

So, on your first page, change
<form action="resultstest.php" methods="post">
to
<form action="resultstest.php" method="post">

And on the second page, change
$categoryID=$GET[categoryID];
to
$categoryID=$POST['categoryID'];

You should always put the quotes in, too (['categoryID'] vs [categoryID]), otherwise you're technically using the constant (if you don't know, a variable that can't be changed and doesn't start with $) instead of the string (word or words) categoryID.

timmybuck

Hi thanks for spotting that silly error i made but still no joy

here is the drop down menu page

<html>

<head>
</head>
<body>
<form action="resultstest.php" method="post">
<?php

mysql_connect('localhost', 'phpuser', 'phpuser03')or die ("Database connection error");

mysql_select_db('booking');

//gets all the available events from the events table.

$event_categories=mysql_query("SELECT * FROM event_categories ")

OR DIE("Unable to find events. Error: ".mysql_error());

print("<select name=\"category\" size=1>\n");

</form>
</body>

</html>

here is the the page i would like to display data on

<?php
$categoryID=$_POST['categoryID'];
$categoryID = addslashes($categoryID);

mysql_connect("localhost","phpuser","phpuser03") or die ("Database connection error");

mysql_select_db('booking');

$sql = "SELECT * FROM event WHERE categoryID='".$_GET['category']."'";
$result = mysql_query($sql);
while ($row = mysql_fetch_array($result)) {
print_r($row);
}

?>
</body>
</html>

shadypalm88

Well, is there anything in the event table for that category ID?

Or, what problem are you having now?

timmybuck

I get a blank page

Well the categoryID field is in my event table next to my eventID field.

The eventID is autoincrementing and the categoryID shows a list of Zeros.

the categoryID in the category table is also incrementing.

does this help at all

shadypalm88

Oh! I see it now. It's using the wrong thing in the query, in 2 ways actually.

So, the line:
$sql = "SELECT FROM event WHERE categoryID='".$_GET['category']."'";
should read:
$sql = "SELECT FROM event WHERE categoryID='$categoryID'";

And I also see on the first page the variable name is category instead of categoryID.

So, the line:
print("<select name=\"category\" size=1>\n");
should read:
print("<select name=\"categoryID\" size=1>\n");

timmybuck

yes there is i choose education form the drop down menu

and in my events table there is an education event next to it

timmybuck

OK we have output, i get everything out of the events table instead of just what i selected in the drop down menu eg education event

first page

print("<select name=\"categoryID\" size=1>\n");

</form>
</body>

</html>

second page

<?php
$categoryID=$_POST['categoryID'];
$categoryID = addslashes($categoryID);

mysql_connect("localhost","phpuser","phpuser03") or die ("Database connection error");

mysql_select_db('booking');

$sql = "SELECT * FROM event WHERE categoryID='$categoryID'";
$result = mysql_query($sql);
while ($row = mysql_fetch_array($result)) {
print_r($row);
}

?>
</body>
</html>

timmybuck

dude u there,

do u think u could help me make the output look more presentable

timmybuck

can anyone help me fix it