gpt4 book ai didi

javascript - 表格没有列 - Google Charts - PHP AJAX

转载 作者:可可西里 更新时间:2023-10-31 23:39:07 26 4
gpt4 key购买 nike

我今天开始学习如何使用 Google Charts,但我有点卡住了。

我有动态数据(每天大约变化 3-4 次)要输入到图表(饼图)中。我使用 AJAX 作为数据源,使用 PHP 作为后端。我尝试这样做但没有成功:

Ajax :

<?php
include $_SERVER['DOCUMENT_ROOT'].'/includes/galaxy-connect.php';

$database = new Connection();
$database = $database->Connect();
$statement = $database->Prepare(" SELECT COUNT(Membership_Level_Name) AS MemTotal, Membership_Level_Name
FROM membership AS M
LEFT JOIN membership_levels AS L
ON M.`Membership_Level_Id` = L.`Membership_Level_Id`
LEFT JOIN membership_status AS S
ON M.`MembershipStatusId` = S.MembershipStatusId
WHERE M.`MembershipStatusId` = 1
GROUP BY L.`Membership_Level_Name`
ORDER BY L.`Membership_Level_Id` ");

$statement->execute();
$MembershipTotals = $statement->fetchall(PDO::FETCH_ASSOC);

if (!empty($MembershipTotals)) {
foreach ($MembershipTotals as $MembershipTotal) {
$data[] = array(
"cols" => array("id"=>"Membership_Level_Name", "label"=>"Membership Level", "type"=>"varchar"),
array("id"=>"MemTotal", "label"=>"Total", "pattern"=>"", "type"=>"number"),
"rows" => array($MembershipTotal['Membership_Level_Name'], $MembershipTotal['MemTotal'])
);
}
}

echo json_encode($data);

这就是我的 ajax,它产生:

(好吧不会让我张贴图片,但这是结果)

[{"cols":{"id":"Membership_Level_Name","label":"Membership Level","type":"varchar"},"0":{"id":"MemTotal","label":"Total","pattern":"","type":"number"},"rows":["Start Up","24"]},{"cols":{"id":"Membership_Level_Name","label":"Membership Level","type":"varchar"},"0":{"id":"MemTotal","label":"Total","pattern":"","type":"number"},"rows":["Member","131"]},{"cols":{"id":"Membership_Level_Name","label":"Membership Level","type":"varchar"},"0":{"id":"MemTotal","label":"Total","pattern":"","type":"number"},"rows":["Member Plus","170"]},{"cols":{"id":"Membership_Level_Name","label":"Membership Level","type":"varchar"},"0":{"id":"MemTotal","label":"Total","pattern":"","type":"number"},"rows":["Premier Member","31"]},{"cols":{"id":"Membership_Level_Name","label":"Membership Level","type":"varchar"},"0":{"id":"MemTotal","label":"Total","pattern":"","type":"number"},"rows":["Bronze","97"]},{"cols":{"id":"Membership_Level_Name","label":"Membership Level","type":"varchar"},"0":{"id":"MemTotal","label":"Total","pattern":"","type":"number"},"rows":["Silver","145"]},{"cols":{"id":"Membership_Level_Name","label":"Membership Level","type":"varchar"},"0":{"id":"MemTotal","label":"Total","pattern":"","type":"number"},"rows":["Gold","188"]},{"cols":{"id":"Membership_Level_Name","label":"Membership Level","type":"varchar"},"0":{"id":"MemTotal","label":"Total","pattern":"","type":"number"},"rows":["Affiliate","3"]},{"cols":{"id":"Membership_Level_Name","label":"Membership Level","type":"varchar"},"0":{"id":"MemTotal","label":"Total","pattern":"","type":"number"},"rows":["Charity\/Education","4"]}]

所以下一步是调用该数据,我从 Google Charts“连接到数据库”(或类似的页面)页面获取了代码:

  <script type="text/javascript" src="https://www.google.com/jsapi"></script>
<script type="text/javascript">

// Load the Visualization API and the piechart package.
google.load('visualization', '1', {'packages':['corechart']});

// Set a callback to run when the Google Visualization API is loaded.
google.setOnLoadCallback(drawChart);

function drawChart() {
var jsonData = $.ajax({
url: "/ajax/charts/membershiptotals.php",
dataType:"json",
async: false
}).responseText;

// Create our data table out of JSON data loaded from server.
var data = new google.visualization.DataTable(jsonData);

// Instantiate and draw our chart, passing in some options.
var chart = new google.visualization.PieChart(document.getElementById('chart_div'));
chart.draw(data, {width: 400, height: 240});
}


</script>

重新加载网页并产生错误:

表格没有列

虽然我不明白为什么..我查看了其他解决方案并在 Quora 和 Google 群组上发布了 API 但无济于事..有人能告诉我代码有什么问题吗?

最佳答案

我找到了答案:

AJAX 更改为:

<?php

include $_SERVER['DOCUMENT_ROOT'].'/includes/galaxy-connect.php';

$database = new Connection();
$database = $database->Connect();
$statement = $database->Prepare(" SELECT COUNT(Membership_Level_Name) AS MemTotal, Membership_Level_Name
FROM membership AS M
LEFT JOIN membership_levels AS L
ON M.`Membership_Level_Id` = L.`Membership_Level_Id`
LEFT JOIN membership_status AS S
ON M.`MembershipStatusId` = S.MembershipStatusId
WHERE M.`MembershipStatusId` = 1
GROUP BY L.`Membership_Level_Name`
ORDER BY L.`Membership_Level_Id` ");
$statement->execute();
$MembershipTotals = $statement->fetchall(PDO::FETCH_OBJ);


$col1=array();
$col1["id"]="";
$col1["label"]="Membership Type";
$col1["pattern"]="";
$col1["type"]="string";

$col2=array();
$col2["id"]="";
$col2["label"]="Total";
$col2["pattern"]="";
$col2["type"]="number";

$cols = array($col1,$col2);

$rows=array();

foreach ($MembershipTotals AS $MembershipTotal) { //foreach ($Event->TrainingTotals['ConfirmedTotal'] AS $Key => $Value) {
$cell0["v"]=$MembershipTotal->Membership_Level_Name;
$cell1["v"]=intval($MembershipTotal->MemTotal);

$row0["c"]=array($cell0,$cell1);
array_push($rows, $row0);
}

$data=array("cols"=>$cols,"rows"=>$rows);
echo json_encode($data);

这使它变得更容易一些,然后在实际页面上:

<script type="text/javascript">
// Load the Visualization API and the piechart package.
google.load('visualization', '1', {'packages':['corechart']});

// Set a callback to run when the Google Visualization API is loaded.
google.setOnLoadCallback(drawChart);
function drawChart() {
var jsonData = $.ajax({
url: "/ajax/charts/membershiptotals.php",
dataType:"json",
async: false
}).responseText;

// Create our data table out of JSON data loaded from server.
var data = new google.visualization.DataTable(jsonData);
var chart = new google.visualization.PieChart(document.getElementById('chart_div'));
chart.draw(data, {title:'Membership Bookings', width: 800, height: 500});
}


</script>

基本上我必须清楚地声明这些列,而 intval 是将它变成一个整数,否则它会将数字作为 Google 不喜欢的字符串返回。希望这对任何人都有帮助:)

感谢 Harish 的回答,但我需要它更有活力 :-)

关于javascript - 表格没有列 - Google Charts - PHP AJAX,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/31746902/

26 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com