Re: [PHP-DB] Odd MySQL Problem...
Its looking for an integer not a string for the number of characters. I think.. Eg : number_pad($n, 4); On Oct 13, 2011, at 9:53 PM, Jason Pruim wrote: number_pad($n, "4"); Best, Karl DeSaulniers Design Drumm http://designdrumm.com
[PHP-DB] Odd MySQL Problem...
Hey everyone,Have a weird issue that I can't seem to figure out...Using PHP to insert phone numbers into the site... I have the Area Code, & Exchange, and then dynamically create the last 4 digits... Once it's inserted it's dropping the leading zero's... Here's some samples:Copied from phpMySQL: Edit Inline Edit Copy Delete2122000681 Edit Inline Edit Copy Delete2122001682 Edit Inline Edit Copy Delete2122002683 Edit Inline Edit Copy Delete2122003684 Edit Inline Edit Copy Delete2122004echoed output from PHP:SQL: INSERT INTO phonesite (areacode, exchange, subscriber) VALUES(212, 200, 0001) SQL: INSERT INTO phonesite (areacode, exchange, subscriber) VALUES(212, 200, 0002) Actual PHP code:// ini_set('display_errors', 1);//error_log(-1);set_time_limit(0);//set_time_limit("120");include "includes.php";include "databaseabstraction.php";include "authentication.php";dbconnect("localhost", "XX", "XX", "XX")or die("Unable to connect: " . mysql_error());function number_pad($number,$n) {return str_pad((int) $number,$n,"0",STR_PAD_LEFT);}if (($handle = fopen("newyorktest.csv", "r")) !== FALSE) { while (($data = "" 1000, ",")) !== FALSE) {print_r($data); foreach( range(0, ) AS $n) { //echo "DATA: " . $data[2] . ""; $padded_number = number_pad($n, "4"); $sql = "INSERT INTO phonesite (areacode, exchange, subscriber) "; $sql .= "VALUES({$data[1]}, {$data[2]}, {$padded_number}) "; //mysql_query($sql) or die("Didn't insert you dumb ass FIX IT NOW CLOWN " . mysql_error());echo "SQL: " . $sql . ""; } } echo "File Bitches!"; fclose($handle);}?>Any ideas on what I'm missing?Thanks everyone! Jason Pruimli...@pruimphotography.com
Re: [PHP-DB] SELECT syntax
Heh, Thanks Karthik. Not my post.. :) But your solution looks dead on.. Here you go Ron. Try this one. Best, Karl On Oct 13, 2011, at 2:42 AM, Karthik S wrote: Try this, select CASE answer when 1 then trivia_answer_1 when 2 then trivia_answer_2 when 3 then trivia_answer_3 when 4 then trivia_answer_4 END as trivia_answers from bible_trivia_table On Thu, Oct 13, 2011 at 1:02 PM, Karl DeSaulniers > wrote: Or something like this? SELECT * FROM `Bible_trivia` WHERE answer=`answer`; Then match the results to trivia_answer_1 in php to see if correct. if($trivia_answer_1 == $results) { ... do this } or a switch switch ($results) { case $trivia_answer_1: ... do this case $trivia_answer_2 ... do this Best, Karl On Oct 12, 2011, at 11:04 PM, Amit Tandon wrote: SELECT `trivia_answer_`answer`` FROM `Bible_trivia` Karl DeSaulniers Design Drumm http://designdrumm.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php Karl DeSaulniers Design Drumm http://designdrumm.com
Re: [PHP-DB] SELECT syntax
Or something like this? SELECT * FROM `Bible_trivia` WHERE answer=`answer`; Then match the results to trivia_answer_1 in php to see if correct. if($trivia_answer_1 == $results) { ... do this } or a switch switch ($results) { case $trivia_answer_1: ... do this case $trivia_answer_2 ... do this Best, Karl On Oct 12, 2011, at 11:04 PM, Amit Tandon wrote: SELECT `trivia_answer_`answer`` FROM `Bible_trivia` Karl DeSaulniers Design Drumm http://designdrumm.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php