This is silly....this code;
if ($debug) { &debug("Wasn't found, so adding $into_cron $query"); }
if ($debug) { &debug("\$query = $query"); }
if ($debug) { &debug("\$into = $into"); }
$back_into_file .= "$into $query\n";
}
Produces;
DEBUG INFO: $query = INSERT INTO
DEBUG INFO: $into = test1234
YET, viewing the contents of $back_into_file shows;
INSERT INTO
...i.e without the $into part! How can the variable exists one line above, but not show on the next? It silly!
Any ideas?
Andy (mod)
andy@ultranerds.co.uk
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Code:
if (!$found) { if ($debug) { &debug("Wasn't found, so adding $into_cron $query"); }
if ($debug) { &debug("\$query = $query"); }
if ($debug) { &debug("\$into = $into"); }
$back_into_file .= "$into $query\n";
}
Produces;
DEBUG INFO: $query = INSERT INTO
DEBUG INFO: $into = test1234
YET, viewing the contents of $back_into_file shows;
INSERT INTO
...i.e without the $into part! How can the variable exists one line above, but not show on the next? It silly!
Any ideas?
Andy (mod)
andy@ultranerds.co.uk
Want to give me something back for my help? Please see my Amazon Wish List
GLinks ULTRA Package | GLinks ULTRA Package PRO
Links SQL Plugins | Website Design and SEO | UltraNerds | ULTRAGLobals Plugin | Pre-Made Template Sets | FREE GLinks Plugins!
Mad...how can it exist, yet it doesnt?