[SOLVED] posting data

[sanchez77]

so I am query one table for a value, and then i want to use that value as part of where clause to update a record in another table. but for some reason that i 

 

what i am doing wrong?

 

//setting value

$masterid =  mysql_query("SELECT masterid FROM lockers WHERE lockerno = '01' LIMIT 0, 30");

//update table with equal value

mysql_query("UPDATE lockersmaster SET timeout = '$timeout', dateout = '$dateout', storetimeout = '$storetimeout' WHERE id = '$masterid'");

[sanchez77]

Can i join them?

[Fadion]

mysql_query() doesn't return a usable value. The code should be:

 

<?php
$results =  mysql_query("SELECT masterid FROM lockers WHERE lockerno = '01' LIMIT 0, 30");
while($values = mysql_fetch_array($results)){ //results from mysql are put in an array and each row is looped
     //set $timeout, $dateout and $storetimeout
     $resultsU = mysql_query("UPDATE lockersmaster SET timeout = '$timeout', dateout = '$dateout', storetimeout = '$storetimeout' WHERE id=" . $values['id']);
?>

[sanchez77]

Should it be

$values['masterid']

?

 

When i tried that, it gave me a blank page and no tables were updated

[ldougherty]

That is exactly correct, it should be..

 

<?php
$results =  mysql_query("SELECT masterid FROM lockers WHERE lockerno = '01' LIMIT 0, 30");
while($values = mysql_fetch_array($results)){ //results from mysql are put in an array and each row is looped
     //set $timeout, $dateout and $storetimeout
     $resultsU = mysql_query("UPDATE lockersmaster SET timeout = '$timeout', dateout = '$dateout', storetimeout = '$storetimeout' WHERE id=" . $values['masterid']);
?>

[sanchez77]

i appreciate your help, but it's not working for, just keeps giving me a blank page, 

[ldougherty]

Try adding this to your code..

 

ini_set('display_errors', 1);

error_reporting(E_ALL);

 

This will show you any errors in your script when executing the page.

 

If it doesn't work just check thoroughly through your code for syntax and try outputting variables before the UPDATE query to see if they are coming through the first query as expected.

 

IE echo $values['masterid']

[Philip]

Or, just in a single query:

UPDATE table1 join table2 on table1.id=table2.id SET table1.some_val = table2.value

 

Which when applied should look something like (untested of course):

UPDATE lockersmaster JOIN lockers ON lockersmaster.id=lockers.masterid SET lockersmaster.timeout = '$timeout', lockersmaster.dateout = '$dateout', lockersmaster.storetimeout = '$storetimeout'

[sanchez77]

that worked. i used it like so:

 

mysql_query("UPDATE lockersmaster JOIN lockers ON lockersmaster.id=lockers.masterid SET lockersmaster.timeout = '$timeout', lockersmaster.dateout = '$dateout', lockersmaster.storetimeout = '$storetimeout'");

 

Cheers,