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จะบันทึกค่าvalue ของ dropdownlist ลงฐานข้อมูลอย่างไรครับ PHP |
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คือผมดึง ข้อมูลจากตารางหนึ่งในDB เพื่อมาทำเป็น dropdownlist อ่าครับ แล้วทีนี้ ผมจะให้มันเก็บค่าทั้งหมดที่เรากรอกลงไปเพิ่มในDBในอีกตารางหนึ่งโดยค่าในdropdownlist ให้เก็บเป็นรหัส ผมจะทำอย่างไรมันจึงจะส่งค่าไปที่อีกหน้าหนึ่งแล้วinsertลงไปในDB ครับ
Code (PHP)
<?php
error_reporting(E_ALL ^ E_NOTICE);
session_start();
$dbhost = "";
$dbname = "";
$dbuser = "";
$dbpass = "";
?>
<Form action="signup.php" method="get">
<TABLE border=1 cellpadding=0 cellspacing=0 bordercolor=white>
<tr><td bordercolor=#660066>
<Table cellspacing=1 bgcolor=#CCFFFF >
<tr>
<td colspan=3 bgcolor=#00CCFF><b> :: Register New Employee</b></td></tr>
<tr><td align=right> Password:</td>
<td><input type="password" name="passwd1" id="passwd1" size=25></td><td></td></tr>
<tr><td align=right>ทวน Password:</td>
<td><input type="password" name="passwd2" id="passwd2" size=25 onBlur="passcheck()"></td><td>
<tr>
<td align=right> ชื่อ- สกุล:</td>
<td><input type="text" name="fname" id="fname" size=25></td><td></td></tr>
<tr><td align=right>Address:</td>
<td><textarea name="msg" id="msg" cols=25 rows=4></textarea></td><td></td></tr>
<tr><td align=right> Telephone:</td>
<td><input id="tel" type="text" name="tel" size=25 onBlur="telcheck()"></td><td>
</td></tr>
<tr>
<td align=right>Fax:</td>
<td><input id="fax" type="text" name="fax" size=25 onBlur="telcheck()"></td><td>
</td></tr>
<tr>
<td align=right> ตำแหน่ง:</td>
<td><?
mysql_connect($dbserver, $dbuser, $dbpass) or die("เชื่อมต่อฐานข้อมูลไม่ได้");
mysql_query("set NAMES UTF8 ");
mysql_select_db($dbname) or die("เลือกฐานข้อมูลไม่ได้"); // เลือกฐานข้อมูล
$sql = "SELECT rank_name from emp_rank Where rank_id ='$rank' ";
$result = mysql_query( $sql );
if (mysql_fetch_array($result) ) {
$current = mysql_result($result, 0, 'rank_name');
} else {
$current = "-กรุณาเลือก-"; $rank = 0;
}
$sql = "SELECT * from emp_rank Order By rank_id ";
$result = mysql_query( $sql);
echo "<SELECT NAME='rank' onChange=\"MM_jumpMenu('parent',this,0)\" >" ;
echo "<option value=$PHP_SELF?rank=$rank>$current</option> \n" ;
echo "<option value=$PHP_SELF?rank=$rank>-------</option> \n" ;
While( $fetcharr = mysql_fetch_array($result) ) {
$rank_id = $fetcharr['rank_id'];
$rank_name = $fetcharr['rank_name'];
echo "<option value=$rank_id > $rank_name </option> \n" ;
}
echo "</SELECT> ";
?></td><td></td></tr>
<tr><td colspan=2 align=right><br>
<input type="image" src="file:///J|/DBMS/images/ok.gif" onClick="register()"> </td><td></td></tr>
</Table>
</td></tr></TABLE>
</form>
Code (PHP)
<?php
error_reporting(E_ALL ^ E_NOTICE);
session_start();
$dbhost = "";
$dbname = "";
$dbuser = "";
$dbpass = "";
$passwd1 = trim($_GET['passwd1']);
$passwd2 = trim($_GET['passwd2']);
$fname = trim($_GET['fname']);
$msg = trim($_GET['msg']);
$tel = trim($_GET['tel']);
$fax = trim($_GET['fax']);
$rank = $_POST['rank_id'];
mysql_connect($dbhost,$dbuser,$dbpass) or die("connect mysql ไม่ได้");
mysql_select_db($dbname);
mysql_query("set NAMES UTF8");
$dbquery = mysql_query($sql);
$sql = "INSERT into emp (emp_pass,emp_name,emp_address,emp_phone,emp_fax,rank_id,emp_cir,branch_id)
values ('$passwd2','$fname','$msg','$tel','$fax','$rank','0','1') ";
$result = mysql_db_query($dbname, $sql);
if (!$result) {
echo("Error in SQL ".mysql_error() );
}
else {
echo "<center>เพิ่มพนักงานเรียบร้อยแล้ว <br/>";
echo
"<br/>".$fname. "<br/>".$msg."<br/>".$tel."<br/>".$rank."<a href=signupform.php>กลับไปหน้าหลัก</a>";
}
}
else {
echo "<b>กรุณากรอกข้อมูลให้ครบถ้วน</b>";
}
?>
Tag : PHP, MySQL
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ประวัติการแก้ไข 2011-01-26 23:37:41
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Date :
2011-01-26 23:35:29 |
By :
guuarmza |
View :
6553 |
Reply :
1 |
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Load balance : Server 05
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