Question

我一直在尝试通过PHP代码将表单数据传递到数据库，但是它不起作用，我已经查看了代码的千分之一，因为可能的错误但是找不到一个作为初学者。表单实际上会提交，但没有任何内容可以访问数据库。

任何快速帮助都将深表感谢。这是代码：

$conn = @mysqli_connect('localhost', 'root', 'aboki'); 
if (mysqli_connect_error()) {
die('Connect Error: ' . mysqli_connect_error());
}

$qry = "INSERT INTO users (email, firstName, surname, userName, password, birthday) values ($email, $firstName, $surname, $userName, $password, $userDOB)";
$result = mysqli_query($conn, $qry);

Answer 1

试试这个

$qry = "INSERT INTO users (email, firstName, surname, userName, password, birthday) 
values ('$email', '$firstName', '$surname', '$userName', '$password', '$userDOB')";

Answer 2

首先，您没有引用未插入的值...

这会解决它（但我强烈建议你不要使用这种方法！）：

$qry = "INSERT INTO users (email, firstName, surname, userName, password, birthday) values ('$email', '$firstName', '$surname', '$userName', '$password', '$userDOB')";

正确的方法

最好充分利用mysqli提供的预定义函数，并在准备好的语句中绑定这些参数，如下所示：

mysqli_prepare($conn,"INSERT INTO users (email, firstName, surname, userName, password, birthday) values (?, ?, ?, ?, ?, ?)");

mysqli_stmt_bind_param($conn, 'TYPES_HERE',$email, $firstName, $surname, $userName, $password, $birthday)

Answer 3

我有数据插入解决方案，你可以尝试一下。

$conn= mysqli_connect("localhost", "root", "my_password", "world");
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
$query = "INSERT INTO users 
(email, firstName, surname, userName, password, birthday) VALUES
($email, $firstName, $surname, $userName, $password, $userDOB)";
 mysqli_query($conn, $query);
 printf ("New Record has id %d.\n", mysqli_insert_id($link));
 mysqli_close($link);

正如你在查询中使用mysqli一样，语法完全不同，

随意提问。

由于

Answer 4

示例：

$stmt = mysqli_prepare($conn, "SELECT District FROM City WHERE Name=?")) {
$stmt->bind_param("s", $city);
$stmt->execute();

无法使用PHP将数据插入MySQL

4 个答案:

正确的方法