通过JavaScript在JavaScript中以时间间隔查询MySQL

Basically I want to query MySQL database using an external PHP script. I want this script to be called every 2 seconds. This 2 seconds interval are initiated with a javascript (jquery, flot), which is as follows:

基本上我想使用外部PHP脚本查询MySQL数据库。我希望每2秒调用一次这个脚本。这个2秒的间隔是用javascript(jquery,flot)启动的,如下所示:

<script type='text/javascript'>
    var data = [];
    var dataset;
    var totalPoints = 50;
    var updateInterval = 1000;
    var now = new Date().getTime();

    function GetData() {
        data.shift();
        while (data.length < totalPoints) {
            var y;

            $.ajax({
                url: 'current_api.php', 
                success: function(currentValue){
                    y = currentValue;
                    console.log(y);
                },
            });

            var temp = [now += updateInterval, y];
            data.push(temp);
        }
    }

    var options = {
        ...
    }

    $(document).ready(function () {
        GetData();

        dataset = [
            { label: "CURRENT READING", data: data }
        ];

        $.plot($("#flot-line-chart"), dataset, options);

        function update() {
            GetData();

            $.plot($("#flot-line-chart"), dataset, options)
            setTimeout(update, updateInterval);
        }

        update();
    });
</script>

Currently, Im getting a NULL value at console.log(y);. My PHP script (current_api.php) that handles MySQL queries is as simple as follows:

目前,我在console.log(y);获得NULL值。我处理MySQL查询的PHP脚本(current_api.php)非常简单如下:

<?php
    require "dbCon.php";    // database credentials
    $databaseName = "MAIN";
    $tableName = "day"

    // OPEN MYSQL DATABASE CONNECTION IN PHP
    $dbs = mysql_select_db($databaseName, $con);

    // FETCH DATA FROM MYSQL DATABASE
    $sql = "SELECT value FROM $tableName ORDER BY id DESC LIMIT 1";
    $result = mysql_query($sql);

    header('Content-Type: application/json');
    while ($row = mysql_fetch_assoc($result)) {
        $currentValue = (int) round($row['value']);
    }
    echo json_encode($currentValue);

    // CLOSE THE DB CONNECTION
    mysql_close($con);
?>

I'm new with AJAX and does not know if what I'm trying to do is possible. Can someone help me to debug why i'm getting a NULL value? Thanks in advance.

我是AJAX的新手,不知道我想做的事情是否可行。有人可以帮我调试为什么我得到一个NULL值?提前致谢。

3 个解决方案

#1

You are calling current_api.php in your ajax script without any data. So there is no query string, no $_GET['dbSelect'] and no database. So your json contains only an undefined variable, NULL.

您在ajax脚本中调用current_api.php而没有任何数据。所以没有查询字符串,没有$ _GET ['dbSelect']而没有数据库。所以你的json只包含一个未定义的变量NULL。

Apart from that this is not correct, you cannot use escaping functions to clean up a user-provided table name, you need to check it against a whitelist.

除此之外,这是不正确的,您不能使用转义函数来清理用户提供的表名,您需要根据白名单进行检查。

#2

Got it!

the problem is the declaration of variable y and the url in $.ajax. All thanks to barmar and jeroen for the hints!

问题是变量y的声明和$ .ajax中的url。一切都归功于barmar和jeroen的提示!

The javascript should be:

javascript应该是:

<script type='text/javascript'>
    var data = [];
    var dataset;
    var totalPoints = 50;
    var updateInterval = 1000;
    var now = new Date().getTime();
    var y;

    function GetData() {
        data.shift();
        while (data.length < totalPoints) {

            $.ajax({
                url: 'current_api.php?dbSelect=R9640E5F1E2', 
                success: function(currentValue) {
                    y = currentValue;
                    console.log(currentValue);
                },
            });

            var temp = [now += updateInterval, y];
            data.push(temp);
        }
    }

    var options = {
        ...
    }

    $(document).ready(function () {
        GetData();

        dataset = [
            { label: "CURRENT READING", data: data }
        ];

        $.plot($("#flot-line-chart"), dataset, options);

        function update() {
            GetData();

            $.plot($("#flot-line-chart"), dataset, options)
            setTimeout(update, updateInterval);
        }

        update();
    });
</script>

where R9640E5F1E2 is the database; and the PHP stays as is. :) Now.. move on to another problem... Query strings inside javascripts.

其中R9640E5F1E2是数据库; PHP保持原样。 :)现在..转到另一个问题...查询javascripts内的字符串。

#3

$.ajax({
      dataType: "json",//insert line: receive data from server to json
      url: 'current_api.php?dbSelect=123', //a method 
      /* a method other
     type:"GET",
     data:{dbSelect:dbSelect},
     url: 'current_api.php',
     */

      success: function(currentValue){
        y = currentValue[0];///edit code   y = currentValue
       console.log(y);
        },
      });