Php 未定义索引：mysqli\u ston

我试图将一些json数据从一个文件解析到一个mysql数据库中。我的原始代码如下：

<?php
$response = array();
$res=array();
$json = file_get_contents('C:\Users\Richard\Desktop\test.json');

if($json!=null){
$decoded=json_decode($json,true);
//$decode= var_dump($decoded);
//$ss=$decode["array"];
//echo $decoded['number'];

if(is_array($decoded["configurationItems"]))
{
foreach($decoded["configurationItems"] as $configurationItems)
//for($i=0;$i>sizeof($decoded["configurationItems"]);$i++)

 {


$configurationItemVersion=$configurationItems["configurationItemVersion"];
echo "<br />","configurationItemVersion:",$configurationItemVersion,"<br    />";

$configurationItemCaptureTime=$configurationItems["configurationItemCaptureTime"];
echo "configurationItemCaptureTime:",$configurationItemCaptureTime,"<br />";

$configurationStateId=$configurationItems["configurationStateId"];
echo "configurationStateId:",$configurationStateId,"<br />";


 $result = mysql_query("INSERT INTO configuration_item(configurationItemVersion,configurationItemCaptureTime,configurationStateId)


VALUES('$configurationItemVersion','$configurationItemCaptureTime','$configurationStateId')")or die("Insert Failed ".mysql_error());;

}// check if row inserted or not
if ($result) {
    // successfully inserted into database
    $response["code"] = 1;
    $response["message"] = "successfully stored";

    // echoing JSON response
    echo json_encode($response);
} else {
    // failed to insert row
    $response["code"] = 2;
    $response["message"] = "Oops! An error occurred.";

    // echoing JSON response
    echo json_encode($response);
}

}
}

---

您必须使用类似的方式连接到mysql

$con=mysqli_connect("localhost","my_user","my_password","my_db");

在这条线路上更换您的电话

$result = mysqli_query($GLOBALS["___mysqli_ston"], "INSERT INTO     configuration_item(configurationItemVersion,configurationItemCaptureTime,configurationStateId) 

用这个

$result = mysqli_query($con, "INSERT INTO     configuration_item(configurationItemVersion,configurationItemCaptureTime,configurationStateId) 

什么错误消息？请详细说明，以便我们能帮助你。另外，由于MySQL注入漏洞，我们还可以查看title@Jelmer我已经用更多的信息修改了这个问题，这可能会帮助你回答我的问题，谢谢
$result = mysqli_query($con, "INSERT INTO     configuration_item(configurationItemVersion,configurationItemCaptureTime,configurationStateId)