如何使用AJAX / JSON使用PHP向MySQL數據庫添加詳細信息？

Question

在這里，我嘗試將兩個variables發送到我的php文件：

我在PHP中不斷收到錯誤mysqli_query() expects parameter 1 to be mysqli and undefined indexes兩個variables mysqli_query() expects parameter 1 to be mysqli and undefined indexes

就像他們沒有正確發送或接收。

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：將variables添加到數據庫

<?php

include("mysqlconnect.php");

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

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

?>

mysqlconnect.php等於：

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

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

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

更新后的版本

<?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();
?>

的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);

我的響應警報根本沒有激活。

Answer 1

您使用mysql_connect，多數民眾贊成在mysqli錯;）看看這里： mysqli

Answer 2

mysqlconnect.php中的mysql_connect返回一個MySQL鏈接標識符，但不返回主php文件中mysqli_query所需的Mysqli對象。 您應該同時使用Mysqli這兩個文件。

Answer 3

您必須使用mysqli_connect（“ myhost”，“ myuser”，“ mypassw”，“ mybd”）代替mysql_connect和mysqli_select_db（connection，dbname）; 代替mysql_select_db（）;