Question

/*php file*/

<?php
require("DBinfo.inc");

$query= "insert into login (first_name,email,password,picture_path) values ('" . $_GET['first_name'] . "'," . $_GET['email'] . "'," . $_GET['password'] . "'," . $_GET['picture_path'] . "')";  

$result = mysqli_query($connect,$query);

if (!$result){
    $output="{'msg':'fail'}";
}else{
    $output="{'msg':'user is added'}";
}

print($output);
mysqli_close($connect);
?>



/*DBinfo.inc*/
<?php


$host="localhost";
$user="root";
$password="root";
$database ="twitter";

$connect = mysqli_connect($host, $user, $password, $database);

if(mysqli_connect_errno()){
    die("Cannot connect to database". mysqli_connect_error());
}

?>



OUTPUT:
{'msg':'fail'}

mysqli正在连接数据库，但未存储数据

我正在使用自己的计算机作为服务器，并运行以下行来执行上述文件：

http://localhost:90/Register.php?first_name=adi&email=a@yagoo.com&password=12345&picture_path=home/u.png

任何人都可以告诉我这是什么错误吗

Answer 1

设置值时，SQL中缺少'。

更改

$query= "insert into login (first_name,email,password,picture_path) values ('" . $_GET['first_name'] . "'," . $_GET['email'] . "'," . $_GET['password'] . "'," . $_GET['picture_path'] . "')";

到

$query= "insert into login (first_name,email,password,picture_path) values ('" . $_GET['first_name'] . "','" . $_GET['email'] . "','" . $_GET['password'] . "','" . $_GET['picture_path'] . "')";

实际上，您最好使用 Prepared Statements 来设置参数，以避免 Sql Injection

mysqli connect命令不起作用

1 个答案: