[SOLVED] storing and using results from a query on another query

[petroz]

Hi Guys,

 

I am having some trouble here... The last query in this script is failing to do what it I have intended.....

 

I would like this query to take the results/$row from the query directly above and then use $row as a variable in the final UPDATE SELECT query.... Any help would be greatly appreciated.

 

Thanks,

P

 

 <?php

$request = $_GET['req'];
$requid = $_GET['requid'];

include "db.php";

$status = '2';

$connection = mysql_connect($host,$user,$password) 
or die ('Couldn?t connect to server.');

$db = mysql_select_db($database,$connection) 
or die ('Cannot Find DB'); 

$introduce = sprintf("SELECT name, organization, email, phone, url, im, address, image FROM users WHERE uid='$requid'",
mysql_real_escape_string($requid));

$contactinfo = mysql_query($introduce);

if (!$contactinfo) {
die("Invalid contactinfo query: " . mysql_error());
}

while($row = mysql_fetch_array($contactinfo))

{

$stringData = $row['name'];
printf($stringData . "«");

$stringData = $row['organization'];
printf($stringData . "«");

$stringData = $row['email'];
printf($stringData . "«");

$stringData = $row['phone'];
printf($stringData . "«");

$stringData = $row['url'];
printf($stringData . "«");

$stringData = $row['im'];
printf($stringData . "«");

$stringData = $row['address'];
printf($stringData . "«");

$stringData = $row['image'];
printf("https://mydomain.com" . $stringData . ".png");

}

$getresp = "SELECT resp_uid FROM requests WHERE request = '$req'";	

$result = mysql_query($getresp) or die(mysql_error());

while($row = mysql_fetch_array($result)){

//print_r($row);

$accept = "UPDATE requests2 SET `status` = '$status', `resp_name` = (SELECT `name` FROM `users` WHERE `uid` = '$row'), `resp_organization` = (SELECT `organization` FROM `users` WHERE `uid` = '$row'), `resp_email` = (SELECT `email` FROM `users` WHERE `uid` = '$row'), `resp_phone` = (SELECT `phone` FROM `users` WHERE `uid` = '$row'), `resp_url` (SELECT `email` FROM `users` WHERE `uid` = '$row'), `resp_im` = (SELECT `im` FROM users WHERE `uid` = '$row'),  `resp_address` = (SELECT `address` FROM `users` WHERE `uid` = '$row')"; 

$acceptresult = mysql_query($accept) or die(mysql_error());

}

?>

[petroz]

Here was my solution...

 

$getresp = "SELECT resp_uid FROM requests WHERE request = '$req'";	

$result = mysql_query($getresp) or die(mysql_error());

while($row = mysql_fetch_array($result)){

$respuid = ($row['resp_uid']);