如何从JavaScript中的MySQL数据库获取数据以构建图表? [英] How to fetch data from MySQL database in JavaScript to build a chart?
问题描述
我正在尝试构建图表，我需要从MySQL数据库表中获取数据到JavaScript变量中，该变量的格式应为:var variable1 = [[1, 19], [2, 11], [3, 14], [4, 16]].我图表上的第一个数字(列)为 x ，第二个数字为 y .我在MySQL数据库中的表看起来像这样(我简化了一点):
I am trying to build a chart and I need to fetch data from the MySQL database table into a JavaScript variable that would be in the following format: var variable1 = [[1, 19], [2, 11], [3, 14], [4, 16]]. The first number(column) becomes x and the second is y on my chart. My table in MySQL database looks like this (I simplified it a bit):
column1 column2
1 19
2 11
3 14
4 16
最简单的方法是什么?我对此并不陌生，请原谅询问可能是一个非常简单的问题.
What is the easiest way to do this? I am new to this, please excuse me for asking what might be a very simple question.
借助Wartus的答案，我编写了如下代码. 我制作了两个文件:带有JavaScript的HTML和一个PHP文件.这是我的HTML文件:
With the help of Wartus' answer I coded as follows. I made two files: HTML with JavaScript and a PHP file. Here is my HTML file:
<html>
<head>
<title>Title</title>
<link rel="stylesheet" type="text/css" href="css/common.css" />
<script language="javascript" type="text/javascript" src="flot/jquery.js"></script> <!-- jQuery library -->
<script language="javascript" type="text/javascript" src="flot/jquery.flot.js"></script> <!-- Library with charts that I plan to use -->
<script type="text/javascript">
$.ajax({
url : 'serv.php', // my php file
type : 'GET', // type of the HTTP request
success : function(result){
var obj = jQuery.parseJSON(result);
console.log(obj);
}
});
</script>
</head>
<body>
Hi
</body>
</html>
这是我的名为 serv.php 的PHP文件，与HTML文件位于同一目录中.
And this is my PHP file named serv.php that is located in the same directory as the HTML file:
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "datadb";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT column1, column2 FROM chartdata"; //This is where I specify what data to query
$result = $conn->query($sql);
echo json_encode($result);
?>
这一切似乎都没有错误，只是我在检查控制台时给出了空值:
It all seems to be working without errors except that it gives out nulls when I check console:
Object {current_field: null, field_count: null, lengths: null, num_rows: null, type: null}
我在做什么错了?
推荐答案
在数据库中进行选择后，您必须以json格式返回答案(对我来说，我只是用要测试的值制作了一个数组) :
After you have make your select in your DB you have to return the answer in json format (for me I have just make an array with the value to test) :
您的php文件(我是serv.php):
Your php file (me is serv.php) :
$data = array([1, 19], [2, 11], [3, 14], [4, 16]);
// replace $data by your code to select in DB
echo json_encode($data);
现在，您必须在javascript代码中获得响应.为此，您必须使用javascript或jQuery(在我的情况下为jQuery)发出"GET"请求:
Now you have to get the response in your javascript code. To do that you have to make a "GET" request in javascript or jQuery (jQuery in my case) :
这是您的js文件:
$.ajax({
url : 'serv.php', // your php file
type : 'GET', // type of the HTTP request
success : function(data){
var obj = jQuery.parseJSON(data);
console.log(obj);
}
});
在obj中，您拥有数据:
所以现在您拥有数据并可以访问，它是一个数组，因此:
So now you have your data and to access, is an array so :
- obj[0] contains [1, 19], obj[0][0] contains 1 and obj[0][1] contains 19
- obj[1] contains [2, 11], obj[1][0] contains 2 and obj[1][1] contains 11 ...
在您的情况下，variable1与obj
修改 使用您的数据库:
在发送答案之前，您必须正确构建数据.因此，在您的情况下，您有一个多维数组，当我将数组推入名为data的数组时，该数组就构成了.
Before to send the answer, you have to build correctly your data. So in your case, you have a multidimensional array, that what i make when I push an array in the array named data.
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "datadb";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT column1, column2 FROM chartdata"; //This is where I specify what data to query
$result = mysqli_query($conn, $sql);
$data = array();
while($enr = mysqli_fetch_assoc($result)){
$a = array($enr['column1'], $enr['column2']);
array_push($data, $a);
}
echo json_encode($data);
这篇关于如何从JavaScript中的MySQL数据库获取数据以构建图表?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
