Phpwho Posted March 20, 2007 Share Posted March 20, 2007 I have this code working, but it just wont print to the screen. My editors can not find any errors either! Help Please!! <?php $PID = $_POST['Personal_ID']; $con = @mysql_connect("cefdata","root",""); if (!$con) { die('Could not connect: ' . @mysql_error()); } $query = "INSERT INTO clockers.clocked (Personal_ID, First_Name, Last_Name, Division) SELECT Personal_ID, First_Name, Last_Name, Division FROM clockers.employees WHERE Personal_ID='{$PID}'"; $results = @mysql_query($query); while ($reg = @mysql_fetch_array($results)) { $clockerid = $reg["Personal_ID"]; $fname = $reg["First_Name"]; $lname = $reg["Last_Name"]; $depart = $reg["Division"]; echo $fname . " " . $lname; echo "<br />"; } @mysql_close($con); ?> Quote Link to comment Share on other sites More sharing options...
DanDaBeginner Posted March 21, 2007 Share Posted March 21, 2007 try to seperate the query.. $query1 = 'for insert' and then for select $query2... Quote Link to comment Share on other sites More sharing options...
Phpwho Posted March 21, 2007 Author Share Posted March 21, 2007 No luck... This way makes the print out but wont write to the database: <?php $PID = $_POST['Personal_ID']; $con = @mysql_connect("cefdata","root",""); if (!$con) { die('Could not connect: ' . @mysql_error()); } $query = "INSERT INTO clockers.clocked (Personal_ID, First_Name, Last_Name, Division) SELECT Personal_ID, First_Name, Last_Name, Division FROM clockers.employees WHERE Personal_ID='{$PID}'"; $yellhi = "SELECT Personal_ID, First_Name, Last_Name, Division FROM clockers.employees WHERE Personal_ID='{$PID}'"; $results = @mysql_query($yellhi); while ($reg = @mysql_fetch_array($results)) { $clockerid = $reg["Personal_ID"]; $fname = $reg["First_Name"]; $lname = $reg["Last_Name"]; $depart = $reg["Division"]; echo $fname . " " . $lname; echo "<br />"; } @mysql_close($con); ?> Quote Link to comment Share on other sites More sharing options...
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