Can't print/echo -- Clueless.

[fuzzy1]

Hey all,
The following script works -- EXCEPT -- I can't get anything to print or echo after the foreach, which tells me there must be something wrong with the query/script, but I haven't a clue what that might be. Anyone???

$active_contacts = "SELECT contacts.id
	FROM contacts, accounts
	WHERE contacts.agency_name=accounts.name
	AND accounts.active='on'
	AND contacts.invalid_email='0'
	AND contacts.email_opt_out !='on'
	AND contacts.deleted='0' ";
$result = @mysql_query ($active_contacts) or die ("Error Retrieving Active Contacts"); // Run the query. 
while ($row = mysql_fetch_assoc($result)) { 
$ActiveContacts[] = $row[id];
}
foreach ($ActiveContacts as $target) {
$campaigns_per_target = "UPDATE contacts AS a INNER JOIN (
SELECT target_id, count(*) AS count
FROM campaign_log
WHERE campaign_log.target_id = '$target'
GROUP BY campaign_log.target_id
) AS a_c ON a.id = a_c.target_id
SET cycle = count WHERE cycle < count"; 
$result = @mysql_query ($campaigns_per_target)or die ("Query failed Updating Contact Cycles ");
}
$affected_rows = mysql_affected_rows (); 
echo $affected_rows;
echo ' - Total Contact Cycles Updated';

[NogDog]

Just on the off chance that errors are getting logged instead of displayed, you might want to add the following at the start of the script:

ini_set('display_errors', 1);
error_reporting(E_ALL);

You can delete that or comment it out once things are working.

If that doesn't tell you anything, you might want to throw in something like the following after your mysql_query() calls to see if any matches are actually being found:

if(mysql_num_rows($result) == 0)
{
   user_error("No rows matched for query: $active_contracts", E_USER_WARNING);
}

(Change $active_contracts to $campaigns_per_target for the query in the foreach loop.)

[fuzzy1]

Thanks NogDog,
That does shed some light on things.

Before, I couldn't find anything in the error logs, now at least I'm getting an error -- or several thousand errors ...
"Notice: Use of undefined constant id - assumed 'id' in ...cycles.php on line 19"

$active_contacts = "SELECT contacts.id
	FROM contacts, accounts
	WHERE contacts.agency_name=accounts.name
	AND accounts.active='on'
	AND contacts.invalid_email='0'
	AND contacts.email_opt_out !='on'
	AND contacts.deleted='0' ";
$result = mysql_query ($active_contacts) or die ("Error Retrieving Active Contacts"); // Run the query. 
while ($row = mysql_fetch_assoc($result)) {
// LINE 19 -->
$ActiveContacts[] = $row[id];

Which I take to mean that there is some confusion over which [id] (contacts.id or accounts.id) is being referenced
however I'm not at all sure how to remedy. Curiously however the script does perform the desired update despite
the apparent ambiguity. Hmmm????

[NogDog]

That notice just means that where you have unquoted string literals as array keys (like $row[id] instead of $row['id']), it rather sloppily assumes that you are using a constant (named 'id' in this example), and since no such constant has yet been defined, it assumes its value to be the same as its name. Most of the time (but not always) you can get away with this, but it's best to use a quoted string literal when that is what you mean. See the "Why is $foo[bar] wrong?" section in the Array Types page for a more complete explanation.

In any case, it would be a good thing to clean them up now (put quotes around those array indexes when they are string literals) just to get those error messages out of the way, leaving it clear as to whether there are any other problems (while also making your code more robust and reliable).

[fuzzy1]

Well DUH!!!! Okay, I fixed that. No biggie right?
Which only leaves me back where I started.

Script Updates db table as desired, (translates to matches found)
Breaks all code that follows.
No errors to be seen or found.
Grrrrrrrrrrrrrr.

I found the following post under "mysql_result" on php.net which
got me to wondering about the syntax of my second $result,
but I couldn't make any sense of the post nor the proposed syntax
and parameters ($query, 0,0) in the context of my particular query.

raz0 at NOSPAM dot worldonline dot dk
23-Aug-2003 10:42
If you want to fetch the result from a mysql query similar to one of these two queries...

$query = mysql_query("SELECT COUNT(*) FROM table");
$query = mysql_query("SELECT LAST_INSERT_ID()");

... you would use mysql_result() like shown below to retrieve the output as an int.

$result = mysql_result($query, 0, 0);