php dropdownlist未定义索引
我是PHP编程的新手,我正在使用下拉列表模块 遇到PHP错误严重性:注意消息:未定义索引:yos 这是密码。我希望有人能帮助我php dropdownlist未定义索引,php,Php,我是PHP编程的新手,我正在使用下拉列表模块 遇到PHP错误严重性:注意消息:未定义索引:yos 这是密码。我希望有人能帮助我 <?php $servername="localhost"; $username="root"; $password=""; $dbname="online_voting_web"; //Create connection $con=mysqli_connect($servername,$username,$password,$dbname); $name
<?php
$servername="localhost";
$username="root";
$password="";
$dbname="online_voting_web";
//Create connection
$con=mysqli_connect($servername,$username,$password,$dbname);
$name = $_POST['firstname'];
$name2 = $_POST['lastname'];
$name3 = $_POST['middlename'];
$username = $_POST['username'];
$pass = md5($_POST['password']);
$yos = isset($_POST['yos']) ? $_POST['yos'] : '';
$id = $_POST['stud_id'];
if(!$name || !$name2 ||$name3 || $id || !$year || !$pass|| !$username){
$error="Please fill empty fields";
include"student_reg.php";
exit();
}
$sql = "SELECT stud_id FROM stud_name WHERE stud_id='$id'";
$query = mysqli_query($con,$sql)or die(mysqli_error($con));
$exist = mysqli_num_rows($query);
if($exist==1){
$nam="<center><h4><font color='#FF0000'>The student ID already exist,register another</h4></center></font>";
unset($id);
include('student_reg.php');
exit();
}
$sql = "SELECT stud_username FROM stud_user WHERE username='".$_POST['username']."'";
$query = mysqli_query($con,$sql)or die(mysqli_error($con));
$exist = mysqli_num_rows($query);
if($exist==1){
$nam="<center><h4><font color='#FF0000'>The student ID already exist,register another</h4></center></font>";
unset($id);
include('student_reg.php');
exit();
}
$sql = "INSERT INTO stud_name(stud_id,studFname,stud_Lname,stud_Mname)
VALUES('$id','$name','$name2','$name3')";
$sql2 = "INSERT INTO stud_course(stud_username,stud_year)
VALUES('$username','$year')";
$sql3 = "INSERT INTO stud_user(stud_id,stud_username)
VALUES('$id','$username')";
$sql4 = "INSERT INTO login_(username,password)
VALUES('$username','$pass')";
$result = mysqli_query($con,$sql);
$result2 = mysqli_query($con,$sql2);
$result3 = mysqli_query($con,$sql3);
$result4 = mysqli_query($con,$sql4);
if(!$result && !$result2 && !$result3 &&!$result4){
die("Error on mysql query".mysqli_error());
}
else{
header("location: index.php");
}
?>
您应该确保$\u POST中存在密钥,然后将其分配给变量:
$yos = isset($_POST['yos']) ? $_POST['yos'] : '';
而且这不是一个错误,只是一条通知消息,您可以跳过它中心,字体?20世纪90年代的whats HTML backSome合理的代码缩进将是一个好主意。它帮助我们阅读代码,更重要的是,它将帮助您为自己的利益调试代码。您可能会被要求在几周/几个月内修改此代码,最后您会感谢我。您的脚本有可能会出现问题。请查看您的HTML中是否有
标记?是的。我的问题解决了D和我遇到了另一个问题haysss。仍然得到了相同的通知:(您在SQL查询中使用了$POST,用变量替换它们(…值(“.”.$POST['stud_id'.]”,“.”.$POST['firstname'..”,“$POST['lastname'.]”,“$POST['username'.]”,“.”,“$POST['course'.]”,“$POST['course'.],“$POST['yos'.]”,“.)OMG现在可以使用这种格式了:值(“$username”,“$pass”)?不,它的pass变量名代替了它的值,你可以这样做:值(“$username”,“$pass”),注意双引号
$yos = isset($_POST['yos']) ? $_POST['yos'] : '';