june_c21

<form id="form1" name="form1" method="post" action=""> Drop Down <select name="select"> <option value="A">A</option> <option value="B">B</option> <option value="C">C</option> </select> </form> when user click on A, it will go to A database. how to write the php code?

hi, i want to make a drop down menu for user to filter data. when user click on it will redirect to generate the filter data from mysql. how to write the code? thanks.

thank you

i re-checked my code and there is no error... can you brieftly explain the details sorry, i'm new in this

hi, i'm stuck in this error code : <?php session_start(); $host = 'localhost'; $user = 'root'; $password = 'admin'; $dbase = 'staff'; $dblink = mysql_connect($host,$user,$password); mysql_select_db($dbase,$dblink); $staff_no = $_POST['staff_no']; $position = $_POST['position']; $date = $_POST['date']; $app_letter = $_POST['app_letter']; $info_form = $_POST['info_form']; $handbook = $_POST['handbook']; $access_card = $_POST['access_card']; $vehicle_sticker = $_POST['vehicle_sticker']; $medical_check_up = $_POST['medical_check_up']; $safety_briefing = $_POST['safety_briefing']; $safety_helmet = $_POST['safety_helmet']; $safety_google = $_POST['safety_google']; $safety_boot = $_POST['safety_boot']; $computer = $_POST['computer']; $tel_ext = $_POST['tel_ext']; $room_no = $_POST['room_no']; $email = $_POST['email']; $stationary = $_POST['stationary']; $uniform = $_POST['uniform']; $tshirt = $_POST['tshirt']; $medical_card = $_POST['medical_card']; $induction_program = $_POST['induction_program']; $query= "INSERT INTO details(staff_no,position,date,app_letter,info_form,handbook,access_card,vehicle_sticker,medical_check_up,safety_briefing,safety_helmet,safety_google,safety_boot,computer,tel_ext,room_no,email,stationary,uniform,tshirt,medical_card,induction_program) VALUES ('$staff_no', '$position','$date','$app_letter','$info_form','$handbook','$access_card','$vehicle_sticker','$medical_check_up','$safety_briefing','$safety_helmet','$safety_google','$safety_boot','$computer','$tel_ext','$room_no','$email','$stationary','$uniform','$tshirt','$medical_card','$induction_program') "; $result = mysql_query($query,$dblink); ?> error : Notice: Undefined index: access_card in C:\Web\staff\add.php on line 17 Notice: Undefined index: medical_check_up in C:\Web\staff\add.php on line 19 Notice: Undefined index: safety_briefing in C:\Web\staff\add.php on line 20 Notice: Undefined index: safety_google in C:\Web\staff\add.php on line 22 Notice: Undefined index: tel_ext in C:\Web\staff\add.php on line 25 Notice: Undefined index: uniform in C:\Web\staff\add.php on line 29 Notice: Undefined index: tshirt in C:\Web\staff\add.php on line 30

ok. i think i get what you mean. thanks again

hi, i got this error... anyone please help. thanks SQL query: CREATE SEQUENCE language_id_seq MySQL said: #1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'SEQUENCE language_id_seq' at line 1

can you explain more clearly?

hi, i need help to solve this problem. Please help. thanks SQL query: CREATE TABLE type_of_title( type_of_title INT CHECK ( type_of_title <6 ) PRIMARY KEY , t_o_t_desc TEXT NOT NULL ); MySQL said: #1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'PRIMARY KEY, t_o_t_desc TEXT NOT NULL )' at line 2

tq. solved

huh? i don't understand. can you briefly explain. sorry, i am new with this

hi, i just can't figure out what's wrong with my code. anyone please help.... thanks CREATE TABLE {$db_prefix}banned ( type tinytext NOT NULL, value tinytext NOT NULL ) TYPE=MyISAM; #1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '{$db_prefix}banned ( type tinytext NOT NULL, value tinytext NOT NULL )' at line 1

try to put $row[0] instead of $$row[0]

Problem with the query: SELECT work_id,unit,title1,fault_rep_id,start_date,end_date,secr_recd,facilitator FROM report WHERE id=f Unknown column 'f' in 'where clause' how come? when i check in database the id is integer

what's wrong ?? $id = $_GET['id']; $query = "SELECT work_id,unit,title1,fault_rep_id,start_date,end_date,secr_recd,facilitator FROM report WHERE id=$id "; $result = mysql_query($query, $dblink); while($myrow = mysql_fetch_row($result)) { $work_id = $myrow[0]; $unit= $myrow[1]; $title1 = $myrow[2]; $fault_rep_id = $myrow[3]; $start_date= $myrow[4]; $end_date = $myrow[5]; $secr_recd=$myrow[6]; $facilitator = $myrow[7]; }Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource

$reg_id = 001; echo "$reg_id"; $reg_id++; i want it printed out as 001 insted of 1.

when i write '001' in the form, the output appear '1' . how to make it become '001' ?

hi, i got problem with this the result suppose to be "reg_no " First, i need to select reg_no from reg_no. then the reg_no will store in databse and the id will ++1. how to write that?

thanks a lot

hi, why i getting this error? i can't figure out what's the problem SQL query: INSERT INTO report( work_id, reg_no, unit, title1, fault_rep_id.start_date, end_date, secr_recd, facilitator ) VALUES ( '$work_id', '$reg_no', '$unit', '$title1', '$fault_rep_id', '$start_date', '$end_date', '$secr_recd', '$facilitator' ) MySQL said: #1136 - Column count doesn't match value count at row 1

i'm using mysql as my database. Table A Staff no Name Details. check for duplicate by details ...

that mean stil need to click on submit button in order to display the name?

example of the code?

that's mean user still need to click on submit button in order for the name to appear? what i want is when user enter their staff no, it automatically generate the name in the same form with staff no.

can you explain more details as i'm new with this. thanks