php - 如何使用 PHP 使用 AJAX/JSON 将详细信息添加到 mySQL 数据库？

Question

这里我尝试将两个变量发送到我的php文件:

我的 PHP 中不断出现错误.. mysqli_query() 期望参数 1 为 mysqli 和两个变量的未定义索引

就像它们没有被正确发送或接收一样。

var name = "John";
var address = "UK";

var sendInfo = {
   Name: name,
   Address: address
};

var params = JSON.stringify(sendInfo);
alert(params);

var httpSend = new XMLHttpRequest();
var php = "http://server/~name/folder/insertOffer.php";
httpSend.open("POST", php, true);

httpSend.onreadystatechange = function()
{
   if(httpSend.readyState == 4 && httpSend.status == 200) {

   }
}

httpSend.send(params);

PHP 文件:将变量添加到数据库

<?php

include("mysqlconnect.php");

$name = $_POST['Name']; 
$address = $_POST['Address'];

mysqli_query($connection,"INSERT INTO offerSelected (Id, Url) VALUES ('".$name."','".$address."')");

?>

mysqlconnect.php 等于:

<?php
$connection = mysql_connect("localhost", "user", "pass");

if(!$connection){
die('Could not connect to server: ' . mysql_error());
}

mysql_select_db("table", $connection);
?>

更新版本

<?php
$connection = mysqli_connect("localhost", "user", "pass", "table");
$stmt = $connection->prepare('INSERT INTO offerSelected (Id, Url) VALUES (?, ?)');
$stmt->bind_param('ss', $name, $address);
$stmt->execute();
?>

JavaScript

var xmlhttp1 = new XMLHttpRequest();
var name = "John";
var address = "UK";
var params = 'Name=' + name + '&Address=' + address;
var php_url = "http://server/~name/folder/insertOffer.php";
xmlhttp1.open('POST', php_url, true);

xmlhttp1.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
xmlhttp1.onreadystatechange = function() {
   if (xmlhttp1.readyState == 4 && xmlhttp1.status == 200) {
       var response1 = xmlhttp1.responseText;
       response1 = JSON.parse(response1);
       alert(response1);
       console.log(response1);
       alert('Check the browser console');
   }
}

xmlhttp1.send(params);

我的回复警报根本没有激活。

Answer 1

你使用mysql_connect，这对mysqli来说是错误的;) 看这里:mysqli