使用xmlhttprequest将JavaScript变量发送到PHP以查询MySQL。不起作用

Question

我正在尝试查询我的MySQL数据库。我的HTML文件中的JavaScript代码是...

<script>
function getGameNumberOfPlays() { 
   var sentValue1 = Number;   
   var xhttp;
   xhttp = new XMLHttpRequest();
   xhttp.onreadystatechange = function() {
      if (this.readyState == 4 && this.status == 200) {
         document.getElementById("gameNumberOfPlays").innerHTML = this.responseText;
      }
   };
   xhttp.open("GET", "numberOfPlays.php", true);
   xhttp.send(sentValue1);
}
</script>

'numberOfplays.php'代码是...

<?php
$dbhost =   'localhost';
$dbuser =   'xxx'; 
$dbpwd  =   'xxx'; 
$dbname =   'xxx';

$conn   =   new mysqli( $dbhost, $dbuser, $dbpwd, $dbname );

$p1 = $_GET['sentValue1'];

$sql= "SELECT COUNT(ID) FROM `data` WHERE `gameNumber` =  ?";

$stmt = $conn->prepare($sql);
$stmt->bind_param('i', $p1);
$stmt->execute();
$result = $stmt->get_result();
$user = $result->fetch_array();
echo "$user[0]";
$conn->close();

?>

上面的代码在我的HTML页面中产生0。

但是，如果我将'numberOfplays.php'的代码更改为...

<?php
$dbhost =   'localhost';
$dbuser =   'xxx'; 
$dbpwd  =   'xxx'; 
$dbname =   'xxx';

$conn   =   new mysqli( $dbhost, $dbuser, $dbpwd, $dbname );

$p1 = $_GET['sentValue1'];

$p2 = 171; // or any number that's in the database

$sql= "SELECT COUNT(ID) FROM `data` WHERE `gameNumber` =  ?";

$stmt = $conn->prepare($sql);
$stmt->bind_param('i', $p2);
$stmt->execute();
$result = $stmt->get_result();
$user = $result->fetch_array();
echo "$user[0]";
$conn->close();

?>

我在HTML页面中获得了正确的数字，这样看来编码还可以，除了如何发送和接收变量'sentValue1'。

如果有人能让我知道如何更改代码，以便MySQL查询与变量“ SentValue1”一起使用，我将非常感激。

数据库中的“ GameNumber”列是这样配置的...

column'gamenumber'gamenumber'gamenumber'

非常感谢。

Answer 1

您没有在XHTTP请求

更改中发送sentValue1：

xhttp.open("GET", "numberOfPlays.php", true);
xhttp.send(sentValue1);

to：

let URL = "numberOfPlays.php?sentValue1=<number>"; 
xhttp.open("GET", URL, true); 
xhttp.send();

or：

xhttp.open("GET", "numberOfPlays.php", true);
xhttp.send(sentValue1=<number>);

您可以检查此 link