我想使用Ajax更新数据，下面是示例代码。SweetAlert被移位，因为它已经更新了，但是它在数据库中没有生效，因此它给出了一个错误。
这是在单击submit按钮时提交表单的Ajax脚本。
<script>
$('#submit').on('click', function(event){
event.preventDefault();
var firstname = $('#firstname').val();
var othername = $('#othername').val();
var gender = $('#gender').val();
var id_type = $('#id_type').val();
var id_number = $('#id_number').val();
var issue_date = $('#issue_date').val();
var business_place = $('#business_place').val();
var food_type = $('#food_type').val();
var screened = $('#screened').val();
var sub_metro = $('#sub_metro').val();
var telephone = $('#telephone').val();
var get_date = $('#get_date').val();
var chit_number = $('#chit_number').val();
var remarks = $('#remarks').val();
var user_id = $('#user_id').val();
var vendor_id = $('#vendor_id').val();
if (firstname!="" &&othername!="" && food_type!=""){
$.ajax({
url: "action/vendor_update.php",
type: "POST",
data: {
firstname:firstname, othername:othername, gender:gender,
id_type:id_type, id_number:id_number, issue_date:issue_date,
business_place:business_place, food_type:food_type, screened:screened,
sub_metro:sub_metro, telephone:telephone, get_date:get_date,
chit_number:chit_number, remarks:remarks, user_id:user_id, vendor_id:vendor_id,
},
cache: false,
success: function(data){
if(data.statusCode=200){
$('#dataForm').find('input:text').val('');
alert('Updated');
})
}
else if (data.statusCode=201)
{
alert('Cannot Update');
}
}
});
}else{
alert('All fields are mandatory');
}
});
这段代码用于php服务器端，用于将数据插入数据库。
<?
session_start();
// include('../includes/session.php');
include('./includes/connection.php');
$query = $dbh->$prepare("UPDATE vendors SET Firstname=:firstname, Othername=:othername, Telephone=:telephone, Gender=:gender, IdType=:id_type, IdNumber=:id_number, IdDate=:issue_date, BusinessPlace=:business_place, FoodType=:food_type, Notes=:remarks, ScreenStatus=:screened, ChitNumber=:chit_number, UpdatedBy=:user_id WHERE VendorId=:vendor_id");
$query->execute([
$firstname = $_POST['firstname'];
$othername = $_POST['othername'];
$telephone = $_POST['telephone'];
$gender = $_POST['gender'];
$id_type = $_POST['id_type'];
$id_number = $_POST['id_number'];
$issue_date = $_POST['issue_date'];
$business_place = $_POST['business_place'];
$food_type = $_POST['food_type'];
$remarks = $_POST['remarks'];
$screened = $_POST['screened'];
$chit_number = $_POST['chit_number'];
$user_id = $_POST['user_id'];
$vendor_id = $_POST['vendor_id'];
]);
// $query->execute($_POST);
if ($query->execute()){
echo json_encode(array("statusCode"=>200));
} else {
echo json_encode(array("statusCode"=>500));
}
?>
下面是经过清理的PHP代码：
<?php
// Tip:
// * Use named placeholders
// * Define the query string inside prepare() so you can't "miss" and run
// the wrong query by accident
// * Use single quotes so you can't interpolate variables by accident
// and create ugly SQL injection bugs
$query = $dbh->prepare('UPDATE vendors SET Firstname=:firstname, Othername=:othername, Telephone=:telephone, Gender=:gender, IdType=:id_type...');
// Use ONE of:
// A) if you have slight differences
$query->execute([
'firstname' => $_POST['firstname'],
'othername' => $_POST['othername'],
'telephone' => $_POST['telephone'],
...
]);
// OR
// B) if you're confident the placeholders match 100% and no clean-up
// such as trim() is necessary.
$query->execute($_POST);
if ($query->execute()) {
echo json_encode(array("statusCode"=>200));
} else {
echo json_encode(array("statusCode"=>500));
}
?>
注意：值得注意的是，像这样的代码不需要存在，任何一个像样的ORM都会让这件事变得很琐碎。这是值得探索的选择，你有那里，因为这可能是如此容易。
