<?php
$servername = "localhost";
$username = "root";
$password = "";
$database = "test";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $database);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
echo "Connected successfully";
$value1=$_POST['txtname'];
$value2=$_POST['cellnumber'];
$value3=$_POST['dist'];
$value4=$_POST['specialization'];
$value5=$_POST['membername'];
$value6=$_POST['date'];
$sql = "INSERT INTO students (StudentName, CellNumber, District, Specialization, PromotionMember, Date)
VALUES ('$value1', '$value2', '$value3', '$value4', '$value5', '$value6')";
if (!mysqli_query($sql)) {
die ('Error: ' . mysql_error());
}
else
{
echo ("معلومات ارایه شده شما ثبت شد");
header("Location: register.html");
}
mysqli_close();
?>
连接成功但第23行的数据插入错误(if(!mysqli_query($ sql)){) enter image description here
答案 0 :(得分:1)
您的查询未运行,因为您正在使用需要2个参数的MySQLi
函数来执行查询。
所以而不是
mysqli_query($sql)
你必须这样做:
mysqli_query($conn, $sql)
您的代码也看起来容易受到SQL注入攻击,因此您想知道如何在MySQLi中转义字符串。我建议你使用prepared语句。 我希望这会有所帮助!
答案 1 :(得分:0)
试试这段代码......
$sql = "INSERT INTO students (StudentName,CellNumber, District, Specialization, PromotionMember, Date)VALUES ('$value1', '$value2', '$value3', '$value4', '$value5', '$value6')";
$q=mysqli_query($conn,$sql);
if (!$q) {
die ('Error: ' . mysqli_error($conn));
}else{header("Location: register.html");}
答案 2 :(得分:0)
尝试此代码..
$sql = "INSERT INTO students (StudentName, CellNumber, District, Specialization, PromotionMember, Date)
VALUES ('$value1', '$value2', '$value3', '$value4', '$value5', '$value6')";
$qw=mysqli_query($conn,$sql);
if($qw)
{
header("Location: register.html");
}