[SOLVED] PHP / MYSQL with links?

[summerpewp]

I'm trying to create a nice help file but running into a few problems. I dont know how to assign a link to the end of my .php?subject=x

 

here is my code...

 

echo '<span class="subh3"><a href="?topicid='.$row["topicid"].'">'.$row['subject'].'</a></span><br />';

 

i want the link to go to ?topicid=account or setup etc.. all of which are distinguished in the mysql database...

 

Right now its coming up with the link ?topicid= and thats it... not going anywhere..

 

Hopefully this makes sense.

 

Thanks for any help.

[pocobueno1388]

You need to put the name of the script before the "?".

 

Instead of

?topicid=

 

It should be

script.php?topicid=

 

Also, you may want to verify that the variable $row['topicid'] isn't empty.

[summerpewp]

i had that before and still no go...

 

i essentially want it to load the info on the right of the links..

 

this code doesn't work for me either, and the row is not empty.

 

echo '<span class="subh3"><a href="index.php?topicid='.$row['topicid'].'">'.$row['subject'].'</a></span><br />';

[pocobueno1388]

What does the URL look like exactly when you click on it? If there is nothing after the topicid=...then that variable is empty.

[Ken2k7]

echo '<span class="subh3"><a href="{$_SERVER['PHP_SELF']}?topicid='.$row['topicid'].'">'.$row['subject'].'</a></span><br />';

[summerpewp]

That above "fix" has an error..

 

the link is empty...

index.php?topicid=

but when i do a manual link to that .. it works fine...

 

so it can't be empty

[Ken2k7]

Can you post more of you code?

[summerpewp]

sure:

 

<?php include('connection.php') ?>

<div class="helpmenu">
<?php 
$query = "SELECT DISTINCT topic FROM help";
$sql1 = mysql_query($query);
while($row = mysql_fetch_array($sql1)){ 
	echo'<h2 class="subh2">',$row[0],'</h2>';

$q2 = "SELECT DISTINCT subject FROM help WHERE topic='$row[0]'";
$rh2 = mysql_query($q2);

while ($row = mysql_fetch_array($rh2)){
	echo '<span class="subh3"><a href="index.php?topicid='.$row['topicid'].'">'.$row['subject'].'</a></span><br />';
		}}



?>
</div>
<div class="helptopic">
<?php 
$query2 = "SELECT * FROM help WHERE topicid = '%s'";
$sprintf = sprintf($query2,mysql_real_escape_string($_GET['topicid']));
$sql2 = mysql_query($sprintf);
while($row = mysql_fetch_array($sql2)){
echo'<div class="question">'.$row['question'].'</div><br />
<div class="answer">'.$row['answer'].'</div><br />';
}

?>
</div>

[pocobueno1388]

You never select "topicid" from the database in your query...that might explain why it's empty.

[summerpewp]

how do i select it and have it do a distinct search as well?

[Ken2k7]

Try

<?php include('connection.php') ?>

<div class="helpmenu">
<?php 
$query = "SELECT DISTINCT topic FROM help";
$sql1 = mysql_query($query);
while($row = mysql_fetch_array($sql1)){ 
	echo'<h2 class="subh2">'.$row[0].'</h2>';

$q2 = "SELECT DISTINCT(topic, topicid) FROM help WHERE topic='{$row[0]}'";
$rh2 = mysql_query($q2);

while ($row = mysql_fetch_array($rh2)){
	echo '<span class="subh3"><a href="index.php?topicid='.$row['topicid'].'">'.$row['subject'].'</a></span><br />';
		}}



?>
</div>
<div class="helptopic">
<?php 
$query2 = "SELECT * FROM help WHERE topicid = '%s'";
$sprintf = sprintf($query2,mysql_real_escape_string($_GET['topicid']));
$sql2 = mysql_query($sprintf);
while($row = mysql_fetch_array($sql2)){
echo'<div class="question">'.$row['question'].'</div><br />
<div class="answer">'.$row['answer'].'</div><br />';
}

?>
</div>

[summerpewp]

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in on line 21

 

dont know what the problem is...

[Ken2k7]

Eh, oh well, worth a shot. Never worked with DISTINCT.

<?php include('connection.php') ?>

<div class="helpmenu">
<?php 
$query = "SELECT DISTINCT topic FROM help";
$sql1 = mysql_query($query);
while($row = mysql_fetch_array($sql1)){ 
	echo'<h2 class="subh2">'.$row[0].'</h2>';

$q2 = "SELECT DISTINCT topicid FROM help WHERE topic='{$row[0]}'";
$rh2 = mysql_query($q2);

while ($row = mysql_fetch_array($rh2)){
	echo '<span class="subh3"><a href="index.php?topicid='.$row['topicid'].'">'.$row['subject'].'</a></span><br />';
		}}



?>
</div>
<div class="helptopic">
<?php 
$query2 = "SELECT * FROM help WHERE topicid = '%s'";
$sprintf = sprintf($query2,mysql_real_escape_string($_GET['topicid']));
$sql2 = mysql_query($sprintf);
while($row = mysql_fetch_array($sql2)){
echo'<div class="question">'.$row['question'].'</div><br />
<div class="answer">'.$row['answer'].'</div><br />';
}

?>
</div>

[summerpewp]

Well i dont want duplicated items from the database to be shown, so i have to do distinct in order to prevent that...

 

I'm rather new to php / mysql which is my problem...

[Ken2k7]

Ah I see, what about this:

<?php include('connection.php') ?>

<div class="helpmenu">
<?php 
$query = "SELECT DISTINCT topic FROM help";
$sql1 = mysql_query($query);
while($row = mysql_fetch_array($sql1)){ 
	echo'<h2 class="subh2">'.$row[0].'</h2>';

$q2 = "SELECT DISTINCT(subject), topicid FROM help WHERE topic='{$row[0]}'";
$rh2 = mysql_query($q2);

while ($row = mysql_fetch_array($rh2)){
	echo '<span class="subh3"><a href="index.php?topicid='.$row['topicid'].'">'.$row['subject'].'</a></span><br />';
		}}



?>
</div>
<div class="helptopic">
<?php 
$query2 = "SELECT * FROM help WHERE topicid = '%s'";
$sprintf = sprintf($query2,mysql_real_escape_string($_GET['topicid']));
$sql2 = mysql_query($sprintf);
while($row = mysql_fetch_array($sql2)){
echo'<div class="question">'.$row['question'].'</div><br />
<div class="answer">'.$row['answer'].'</div><br />';
}

?>
</div>

[summerpewp]

Thanks!! That worked perfectly