如何將數組發送到PHP文件，以便可以使用HTTP請求將值添加到mySQL？

Question

將javaScript數組發送到PHP文件，然后將元素存儲在mySQL數據庫中。

目前，我的“ httpSend.responseText”警報出現錯誤。 聲明：未定義索引：..行中的名稱8

注意：未定義的索引：第9行中..中的地址

警告：mysqli_query（）期望參數1為mysqli，第12行在..中給定的資源

我不確定該數組是正確發送還是被正確接收。

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);

的PHP

 <?php

 include("mysqlconnect.php");


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


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

mysqlconnect看起來像這樣

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

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

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

Answer 1

嘗試這個 ：

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代碼：

<?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

為什么使用單引號將變量“ $ name”和“ $ address”包裝起來，將代碼更改為此，這可能會幫助您：

<?php

 include("mysqlconnect.php");


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


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

Answer 3

嘗試這個：

<?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在哪里定義？