Andu Jame Posted May 22, 2007 Share Posted May 22, 2007 can any1 help me?? Is that any error on this php code?? $ip1 = getenv('REMOTE_ADDR'); $q="select * from kawalan_punch where id_kursus ='$id_kursus'"; $row=mysql_fetch_array($r); $ip2=$row["ip_client"]; Link to comment https://forums.phpfreaks.com/topic/52419-can-use-this-code/ Share on other sites More sharing options...
btherl Posted May 22, 2007 Share Posted May 22, 2007 $ip1 = getenv('REMOTE_ADDR'); $q="select * from kawalan_punch where id_kursus ='$id_kursus'"; $are = mysql_query($q) or die("Error in $q\n" . mysql_error()); $row=mysql_fetch_array($are); $ip2=$row["ip_client"]; I added a missing line. Where does $id_kursus come from? What do you want the code to do? Link to comment https://forums.phpfreaks.com/topic/52419-can-use-this-code/#findComment-258694 Share on other sites More sharing options...
Andu Jame Posted May 22, 2007 Author Share Posted May 22, 2007 kawalan_punch is my db and $id_kursus is one of the row. $id_kursus is id for my workers....and i'm doing punch card system for attendance i want this code to verify according to $id_kursus. If $id_kursus is different it will show error msg Link to comment https://forums.phpfreaks.com/topic/52419-can-use-this-code/#findComment-258700 Share on other sites More sharing options...
trq Posted May 22, 2007 Share Posted May 22, 2007 If $id_kursus is different it will show error msg Then you'll need some sort of check. <?php $q = "select * from kawalan_punch where id_kursus ='$id_kursus'"; if ($are = mysql_query($q)) { if (!mysql_num_rows($are)) { echo "id not found"; } else { $row = mysql_fetch_array($are); $ip2 = $row["ip_client"]; } } else { echo "Error in $q\n" . mysql_error(); } ?> Link to comment https://forums.phpfreaks.com/topic/52419-can-use-this-code/#findComment-258701 Share on other sites More sharing options...
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