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Posts posted by yuttakarn
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error_reporting(0);
That's not a good idea. If errors happen you should fix them, not hide them.include("connect.inc.php");
You include that file twice, it only needs to be included once.if (isset($_POST['submit'])) { $province_id = $_POST['province_id'][$i]; $province_id = $_POST['travel_id'][$i]; $i = 0; foreach ( $_POST as $val) { mysql_query("INSERT INTO travel_list (province_id, travel_id) VALUES ('$province_id', '$travel_id')"); $i++; } }
You're trying to use a variable called $travel_id in the query, but you don't have that variable. You saved both values as $province_id. You also don't want to loop over all of $_POST, if you have an array of values then you loop over the array inside $_POST. But your form doesn't have multiple values for each of those, you only have 2 select elements and both of them only let you select 1 value each. So you don't need to loop at all, you can just get the selected values and insert them.Also, your code isn't going to work in the current version of PHP, because functions like mysql_query were removed. Instead, you should use the mysqli extension, or PDO. Both of those support prepared statements, which you should also use.
Please help me I solved this does not come for another three days.
Please help me Project BA
Time remaining 1 day
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Hi I want to bring this information to insert multiple records from select
<?PHP error_reporting(0); include("connect.inc.php"); ?> <?php include "connect.inc.php"; if (isset($_POST['submit'])) { $province_id = $_POST['province_id'][$i]; $province_id = $_POST['travel_id'][$i]; $i = 0; foreach ( $_POST as $val) { mysql_query("INSERT INTO travel_list (province_id, travel_id) VALUES ('$province_id', '$travel_id')"); $i++; } } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <link href="css/bootstrap.min.css" rel="stylesheet" media="screen"> <script src="js/bootstrap.min.js"></script> <title>Untitled Document</title> </head> <body> <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.2/jquery.min.js"></script> <script src="respond.js"></script> <form action="index.php" method="post"> <!--ส่งค่า post ไปหน้าเดิม --> <table width="896" border="1"> <thead> <tr> <th width="239">จังหวัด</th> <th width="552">สถานที่ท่องเที่ยว</th> <th width="83"><input type="hidden" id="txtNum" value="1" size="2" /></th> </tr> <tr> <th width="239"> <select id="selProvince" name="province_id[]"> <!--Default จังหวััด--> <option value="">กรุณาเลือกจังหวัด</option> <?PHP $SelectPr="SELECT * FROM province"; $QueryPro=mysql_query($SelectPr); while($Pro=mysql_fetch_array($QueryPro)){ ?> <option value="<?=$Pro['province_id']?>"><?=$Pro['province_name']?></option> <?PHP } ?> </select> </th> <th width="552"><select name ="travel_id[]" id="selTravel"><option value="">กรุณาเลือกจังหวัด</option></select></th> <th width="83"><button type="button" id="btnP">เพิ่มรายการ</button></th> </tr> <tr><td colspan="3"><center>รายการที่เพิ่ม</center></td></tr> </thead> <tbody> </tbody> </table> <input name="submit" type="submit" value="add"> </form> <br><br><br> <!--ทอสอบ ค่าแสดงผล Muti Atrray--> </body> </html>
I'm insert 1 record but I choose just one , but more data is imported into fifth data .
Informationprovince_id, travel_id not insert into
HELP ME HOT FIX CODE
I choose just one , but more data is imported into fifth data .
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Help ? I want my site to support the mobile.
I must add that anythank you
For use if else check the availability
in PHP
Posted
Customers pick When finished If on the book, such as the selection of Van Dyke VT01 from 1/07/59 - 2/07/59-day period, you will not be able to book your car again. It is based on the availability of the vehicle.
I made a separate table Storage booking Van table separately too.
Tables, storage, booked by id_car I take from the table and put the van.
Table of van
This is the code that I defined it booked only three people per day.
Sorry I am not good at English Please help me