如何将数组发送到 PHP 文件，以便可以使用 HTTP 请求将值添加到 mySQL？

Question

将 javaScript 数组发送到 PHP 文件，然后将元素存储在 mySQL 数据库中。

目前，我的“httpSend.responseText”警报出现错误。说注意：未定义的索引：名称在..第8行

注意：未定义的索引：第 9 行 .. 中的地址

警告：mysqli_query() 期望参数 1 为 mysqli，资源在第 12 行的 .. 中给出

我不确定阵列是正确发送还是被正确接收。

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

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

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

httpSend.onreadystatechange = function() {//Call a function when the state changes.
    if(httpSend.readyState == 4 && httpSend.status == 200) {
        alert(httpSend.responseText);
    }
}

httpSend.send(sendInfo);

PHP

 <?php

 include("mysqlconnect.php");


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


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

mysqlconnect 是这样的

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

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

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

Answer 1

试试这个：

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

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

var params = "sendInfo=" + JSON.stringify(sendInfo);

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

httpSend.onreadystatechange = function() {//Call a function when the state changes.
    if(httpSend.readyState == 4 && httpSend.status == 200) {
        alert(httpSend.responseText);
    }
}

httpSend.send(params);

PHP 代码：

<?php

 include("mysqlconnect.php");

 $sendInfo = json_decode($_POST['sendInfo']);

 $name = $sendInfo ['name']; 
 $address = $sendInfo ['address'];


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

Answer 2

为什么要使用单引号来包装变量 '$name' 和 '$address' 将您的代码更改为这可能会对您有所帮助：

<?php

 include("mysqlconnect.php");


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


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

Answer 3

试试这个：

<?php

    include("mysqlconnect.php");

    $name = $_POST['Name']; // NOTE THE CASE CHANGE HERE AS THIS IS WHATS DEFINED IN YOUR JS
    $address = $_POST['Address'];

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

?>

$connection 是在哪里定义的？