Question

我试图使用jQuery向mysql服务器提交表单。我遇到了jQuery验证问题（当然）。

编辑：发布相关的PHP代码，显然，我的HTML和jQuery没有导致我的错误。

<body>
<?php
//establish connection
$con = new MySQLi ("localhost", "root", "pwdpwd", "pet_shop");

//check connection
if(mysqli_connect_errno())
{
    echo 'Could not connect to server!' .mysqli_connect_error();
}

//Escape variables for security. Use the name of each input item.
$fname = mysqli_real_escape_string($con, $_POST['fname']);
$lname = mysqli_real_escape_string($con, $_POST['lname']);
$address = mysqli_real_escape_string($con, $_POST['address']);
$state = mysqli_real_escape_string($con, $_POST['state']);
$city = mysqli_real_escape_string($con, $_POST['city']);
$zipcode = mysqli_real_escape_string($con, $_POST['zipcode']);
$phone = mysqli_real_escape_string($con, $_POST['phone']);
$petname = mysqli_real_escape_string($con, $_POST['petname']);
$apptDetails = mysqli_real_escape_string($con, $_POST['apptDetails']);
$email = mysqli_real_escape_string($con, $_POST['email']);
$breed = mysqli_real_escape_string($con, $_POST['breed']);
$petType = mysqli_real_escape_string($con, $_POST['petType']);
$neut = mysqli_real_escape_string($con, $_POST['neut']);
$petBday = mysqli_real_escape_string($con, $_POST['petBday']);

$sql = "INSERT INTO grooming (FirstName, LastName, Address, City, State, Zip, PhoneNumber, Email, PetType, Breed, PetName, NeuteredOrSpayed, PetBirthday, AppointmentDetails)
VALUES ('$fname', '$lname', '$address', '$state', '$city', '$zipcode', '$phone', '$petname', '$apptDetails', '$email', '$breed', '$petType', '$neut', '$petBday')";

//check for errors
if (!mysqli_query($con,$sql)) {
    die('Error: ' . mysqli_error($con));
}
echo "new record added";

//close connection
mysqli_close($con);

?>
</body>
</html>

当我点击提交时会发生的事情是输入Pet的名字通过约会详细信息提交到错误的sql列。看起来like this。

我认为这个错误正在发生，因为我忘了在一些输入（品种，宠物类型）中添加正确的类，但是在我尝试之后发生了同样的错误。我不知所措;要么我丢失了代码，要么我误解了代码应该做的事情。任何帮助将不胜感激，谢谢！

Answer 1

在您的插入中查询您接受的内容并将其他内容传递给Values()

更改此行

INSERT INTO grooming (FirstName, LastName, Address, City, State, Zip, PhoneNumber, Email, PetType, Breed, PetName, NeuteredOrSpayed, PetBirthday, AppointmentDetails)
VALUES ('$fname', '$lname', '$address', '$state', '$city', '$zipcode', '$phone', '$petname', '$apptDetails', '$email', '$breed', '$petType', '$neut', '$petBday')";

到这个

INSERT INTO grooming (FirstName, LastName, Address, City, State, Zip, PhoneNumber, Email, PetType, Breed, PetName, NeuteredOrSpayed, PetBirthday, AppointmentDetails)
VALUES ('$fname', '$lname', '$address', '$city','$state', '$zipcode', '$phone', '$email', '$petType', '$breed', '$petname', '$neut', '$petBday', '$apptDetails')";

值通过PHP提交到错误的SQL列

1 个答案: