why doesn't my query work - php

Question

My query stop working suddently. I know it is dutch but can you detect any errors?
There are no MySQL errors. He just won't write $content to the screen.
I tried numerous things. he just stop working
<?php
$activiteiten= "SELECT
activiteiten.ActiviteitID,
activiteiten.Naam,
activiteiten.plaatsen,
agenda.TijdBegin,
agenda.TijdEind,
agenda.AgendaID,
reserveren.ReserveringTijd
FROM
activiteiten,
agenda,
reserveren
WHERE
agenda.ActiviteitID = activiteiten.ActiviteitID
AND
agenda.AgendaID = reserveren.AgendaID
";
$result = mysql_query($activiteiten) or die(mysql_error());
$content .= '<tr>';
$content .= '<td>'.$record['Naam'].'</td>';
$content .= '<td>'.$record['ReserveringTijd'].'</td>';
$content .='<td>Beschikbare plaatsen:'.$record['plaatsen'].'</td>';
$agendaID= $record['AgendaID'];
$agendaData = agendaData($agendaID);
$content.= '<td>aantal reserveringen:'.$agendaData["reserveringen"].'</td>';
$content.= '</tr>';
}

Answer 1

You do a mysql_query, but you never fill the $record variable. How do you expect it to have content?
Try adding:
$record = mysql_fetch_array( $result );
after the mysql_query.
You need to understand these are two different things:
mysql_query will execute a query on the server
mysql_fetch_array (and friends) will get the results in the desired format.
If you only do 1, you execute the query. That works fine, but if you are interested in the results, you need to fetch them. Sometimes you don't need to (for example, when the query is a not a select).

Answer 2

There are no MySQL errors. He just won't write $content to the screen.
What about adding:
echo $content;
If you don't print it, then you won't see anything :)
Apart from that, aren't you mixing $result and $record?

Answer 3

Your code is missing mysql_fetch_assoc call after mysql_query, it should be something like this:
$result = mysql_query($activiteiten) or die(mysql_error());
while(false != ($record = mysql_fetch_assoc($result))
{
$content .= '<tr>';
$content .= '<td>'.$record['Naam'].'</td>';
$content .= '<td>'.$record['ReserveringTijd'].'</td>';
$content .='<td>Beschikbare plaatsen:'.$record['plaatsen'].'</td>';
$agendaID= $record['AgendaID'];
$agendaData = agendaData($agendaID);
$content.= '<td>aantal reserveringen:'.$agendaData["reserveringen"].'</td>';
$content.= '</tr>';
}

Answer 4

Stopped working and not printing out any error rises the suspicion that the data in the database has somehow changed and the query simply does not get any rows.
Execute your query in a database environment like phpMyAdmin and check if any rows are returned.
Otherwise check the other answers for mysql-php-related errors.

Answer 5

$result = mysql_query($activiteiten) or die(mysql_error());
while ($record = mysql_fetch_array($result, MYSQL_ASSOC))
{
// Your $record now contains the data you want
}

Answer 6

You have the query result stored in $result but you do not then go on to do anything with $result - you need to do something like:
$record = mysql_fetch_assoc($qry_result);
to retrieve the record results from $result. If you expect more than one result you'll need to use a while loop to loop through all the results

Answer 7

Change:
$content .= '<tr>';
To:
$content = '<tr>';
And remember to echo the $content on the page.
Make sure you have error reporting on so you see immediately if the problem lies in uninitialized PHP variable or somewhere else.