Trying to run 2nd query from returned data in table??

Helaman

I am a week old on php and need a little help, I am sure this one is easy for an experienced programmer.

my page runs a query and returns the result back into a html table. That part works and is easy learned from many examples in books.

now I want to pull data (id) fom one column and search on that and return it also back as a html table. I can't figure out how to carry the data forward and run a search on an "array?" of data.

my first query (stripped out html )

$query = "SELECT CONCAT(firstname, ' , ' , lastname) AS name, email, id FROM maintable where referrer = '$memid' ORDER by (id) ASC";
$result = @ ($query);//Run the query.
if ($result)
{
while ($row = mysql_fetch_array($result)){
$row[0]</td><td><a href=\"mailto:
$row[1]\">$row[1]</a></td><td>
$row[2]</td></tr>\n";
}

now I want to use $row[2] (user's id) and search again finding the members who have been referrered from this list of members. Of course, I understand $row[2] is now containing a list of different values, so how do I run a query on $row[2]??

is is a basic matrix - for a member to see who is on their first line and then who is on their second line, naturally, the first line is referring the second line.

thanks for any clues on this

Helaman

I am trying running a query again on just selecting the id, then using a while statement to run another query using those results, but of course not working.
I am not sure whether to convert the results in a string or an array?

dalecosp

Try not to worry so much about presentation (HTML, tables, etc.) until your code does what you want. Have you read "Debugging 101" (see linkage in my .sig) ...

Helaman

ok, I am not worried about the html, just how to use the returned data to do another query and present that data

khendar

So - basically this is what you want to do.

Get the name, email address and id for a member.
Get the name, email address and id for members who were referred by the above member.

This is what your code is doing:

$query = "SELECT CONCAT(firstname, ' , ' , lastname) AS name, email, id FROM maintable where referrer = '$memid' ORDER by (id) ASC"; // Getting a list of members
$result = @mysql_query ($query);//Run the query.
if ($result)
{
while ($row = mysql_fetch_array($result)){ // loop through the results

$row[0]</td><td><a href=\"mailto: // print name
$row[1]\">$row[1]</a></td><td> // print email
$row[2]</td></tr>\n"; // print id

}

So each time the while loop is run - you are resetting $row to be the next member in the list of results. $row[2] is not a list - its is a single variable - containing the id of the current member.

So if you want to get the list of members referred by the current member, you do it inside the while loop:

while ($row = mysql_fetch_array($result)){ // loop through the results
echo  $row[0];
echo "<a href=\"mailto:$row[1]\">$row[1]</a>";
echo $row[2];

$curr_id = $row[2]; // put current id into a variable 

$query = "SELECT CONCAT(firstname, ' , ' , lastname) AS name, email, id FROM maintable where referrer = '$curr_id' ORDER by (id) ASC"; 

// Do similar query processing and results presentation

}

Make sense?

Helaman

it makes sense, but when I run it I get,

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/weeklytr/public_html/pages/userlogged.php on line 49

and

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/weeklytr/public_html/pages/userlogged.php on line 37

those are my two while statement lines. also only get result on the first id of the first line. everything else is blank except for those error messages.

khendar

Well that probably means the query fails somewhere along the track - try some error handling/debugging:

$result = mysql_query($query) or die ("Error:" . mysql_error());

You should probably make sure that the second query is named differently eg $inner_result - mysql_query() - otherwise the outer while loop might get confused.

Helaman

thanks got it

BTW, how do you make your example code go into a box, I have been searching all the help files here and don't see where it explains how to insert code into a php code box?? Must be an inside knowledge kind of thing.

I would like to share how I finally got my problem resolved but would like to use one of those boxes.

Helaman

dalecosp

Well, it's a secret if you're using "quick reply" at the page bottom, but, if you "go advanced" you can click the "PHP" icon (looks like a piece of paper with rust colored lettering) and you'll get a BBcode tag.

The secret is [PHP] <?php //your code here... ?> [/PHP] 😉