我是.js文件及其使用的新手,我正在尝试更新使用JS的图表。我需要或尝试做的是从我的数据库导入信息,以便在我的JS文件中使用它来填充图表。这是图表代码和文件名。
文件名charjs_custom.js
/*Polar chart*/
var polarElem = document.getElementById("polarChart");
var data3 = {
datasets: [{
data: [
20,
16,
7,
3,
40
],
backgroundColor: [
"#7E81CB",
"#1ABC9C",
"#B8EDF0",
"#B4C1D7",
"#01C0C8"
],
hoverBackgroundColor: [
"#a1a4ec",
"#2adab7",
"#a7e7ea",
"#a5b0c3",
"#10e6ef"
],
label: 'My dataset' // for legend
}],
labels: [
"Blue",
"Green",
"Light Blue",
"grey",
"Sea Green"
]
};
new Chart(polarElem, {
data: data3,
type: 'polarArea',
options: {
elements: {
arc: {
borderColor: ""
}
}
}
});
我需要使用来自我的数据库的信息来改变“数据”和“标签”部分,从我所看到的我需要创建一个php文件来检索信息但是我如何将其转换为JS以及如何判断它在JS文件中使用什么信息。还链接“数据”和“标签”部分,以便信息对应于我的表
我想使用的表是:make:id~count
档案名称chart_test.php
<?php
//database
$host="my_host"; // Host name
$username="my_username"; // Mysql username
$password="my_password"; // Mysql password
$db_name="my_database"; // Database name
//get connection
$mysqli = new mysqli($host, $username, $password, $db_name);
if(!$mysqli){
die("Connection failed: " . $mysqli->error);
}
//query to get data from the table
$query = sprintf("SELECT * FROM bureau GROUP BY make ");
//execute query
$result = $mysqli->query($query);
//close connection
$mysqli->close();
//now print the data
print json_encode($data);
?>
答案 0 :(得分:0)
您可以使用JSON格式在php后端和js前端之间交换数据。但是我建议你使用nodejs作为后端服务。它将与您的js代码无缝集成。
答案 1 :(得分:0)
如果不了解您的数据源,通常会像以下一样工作。
<?php
// Lets say your managed to convert your data from your database into something like.
$graph = [
'data' => [20, 16, 7, 3, 40],
'background' => ['#7E81CB', '#1ABC9C', '#B8EDF0', '#B4C1D7', '#01C0C8'],
'hover' => ['#a1a4ec', '#2adab7', '#a7e7ea', '#a5b0c3', '#10e6ef'],
'labels' => ['Blue', 'Green', 'Light Blue', 'Grey', 'Sea Green']
];
?>
var polarElem = document.getElementById("polarChart");
var data3 = {
datasets: [{
data: <?php echo json_encode($graph['data']); ?>,
backgroundColor: <?php echo json_encode($graph['background']); ?>,
hoverBackgroundColor: <?php echo json_encode($graph['hover']); ?>,
label: 'My dataset'
}],
labels: <?php echo json_encode($graph['labels']); ?>
};
答案 2 :(得分:0)
因此,如果我理解数据库的映射,我会这样做:
<?php
//database
$host="my_host"; // Host name
$username="my_username"; // Mysql username
$password="my_password"; // Mysql password
$db_name="my_database"; // Database name
//get connection
$mysqli = new mysqli($host, $username, $password, $db_name);
if(!$mysqli){
die("Connection failed: " . $mysqli->error);
}
//query to get data from the table
$query = sprintf("SELECT * FROM bureau GROUP BY make ");
//execute query
$result = $mysqli->query($query);
// Define configuration array
$config = array(
'datasets' => array(
array(
'data' => array(),
'backgroundColor' => array(), // This could be statically loaded, or dynamic if the DB has colors
'hoverBackgroundColor' => array(), // This could be statically loaded, or dynamic if the DB has colors
'label' => 'My Dataset'
)
),
'labels' => array()
);
// Loop through database result and add
while($row=mysqli_fetch_assoc($result)){
array_push($config['labels'], $row['make']); // Add label
array_push($config['datasets'][0]['data'], $row['totalValue']); // Add value
}
//close connection
$mysqli->close();
//now print the data
echo json_encode($config);
您最终使用该配置执行的操作取决于您加载此图表的方式。这可能是一个AJAX请求,然后您将使用结果代替data3
,例如:
$.get( "chart_test.php", function( data ) {
new Chart(polarElem, {
data: data,
type: 'polarArea',
options: {
elements: {
arc: {
borderColor: ""
}
}
}
});
}, "json" );
或者嵌入到php文件中,这样就可以创建整个脚本而不是打印json_encode($config);
:
new Chart(polarElem, {
data: <?= json_encode($config) ?>,
type: 'polarArea',
options: {
elements: {
arc: {
borderColor: ""
}
}
}
});
这一切都取决于你到目前为止的设置。