使用ajax调用php页面不起作用
我在使用ajax调用php页面时遇到问题。单击按钮时,不会显示chart.php。这两个按钮都不起作用。目标php文件是相同的,只是数据不同,所以我只包含了一个chart.php页面。我怎样才能解决这个问题?请谢谢你的帮助 Index.php代码:使用ajax调用php页面不起作用,php,ajax,Php,Ajax,我在使用ajax调用php页面时遇到问题。单击按钮时,不会显示chart.php。这两个按钮都不起作用。目标php文件是相同的,只是数据不同,所以我只包含了一个chart.php页面。我怎样才能解决这个问题?请谢谢你的帮助 Index.php代码: <!DOCTYPE html> <html> <head> <title>Charts</title> <link rel="stylesheet" type="tex
<!DOCTYPE html>
<html>
<head>
<title>Charts</title>
<link rel="stylesheet" type="text/css" href="etc/main.css" />
<script type="text/javascript" src="etc/jquery-2.1.4.js"></script>
</head>
<body>
<div id="wrapper">
<div id="header">
<button id='actbutton1'> Chart1</button>
<button id='actbutton2'> Chart2</button>
</div>
<div id="contentliquid">
<div id="content">
<div id="querydiv">
</div>
</div>
</div>
<div id="leftcolumn">
</div>
</div>
<script>
document.getElementById('actbutton1').onclick = function() {
var xhr = new XMLHttpRequest();
xhr.open("GET", "chart1.php", true);
xhr.onreadystatechange = function() {
if (xhr.readyState == 4 && xhr.status == 200) {
document.getElementById("querydiv").innerHTML = xhr.responseText;
}
}
xhr.send();
}
document.getElementById('actbutton2').onclick = function() {
var xhr = new XMLHttpRequest();
xhr.open("GET", "chart2.php", true);
xhr.onreadystatechange = function() {
if (xhr.readyState == 4 && xhr.status == 200) {
document.getElementById("querydiv").innerHTML = xhr.responseText;
}
}
xhr.send();
}
</script>
</body>
</html>
图表
图表1
图表2
document.getElementById('actbutton1')。onclick=function(){
var xhr=new XMLHttpRequest();
open(“GET”,“chart1.php”,true);
xhr.onreadystatechange=函数(){
如果(xhr.readyState==4&&xhr.status==200){
document.getElementById(“querydiv”).innerHTML=xhr.responseText;
}
}
xhr.send();
}
document.getElementById('actbutton2')。onclick=function(){
var xhr=new XMLHttpRequest();
open(“GET”,“chart2.php”,true);
xhr.onreadystatechange=函数(){
如果(xhr.readyState==4&&xhr.status==200){
document.getElementById(“querydiv”).innerHTML=xhr.responseText;
}
}
xhr.send();
}
我用ajax调用的页面chart1.php如下:
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "mydb";
$con = new mysqli($servername, $username, $password, $dbname);
if ($con->connect_error) {
die("Connection failed: " . $con->connect_error);
}
else
{
}
$query = "SELECT Week, Omean FROM charts";
$aresult = $con->query($query);
?>
<!DOCTYPE html>
<html>
<head>
<title>Chart1</title>
<script type="text/javascript" src="loader.js"></script>
<script type="text/javascript">
google.charts.load('current', {'packages':['corechart']});
google.charts.setOnLoadCallback(drawChart);
function drawChart(){
var data = new google.visualization.DataTable();
var data = google.visualization.arrayToDataTable([
['Week','Omean'],
<?php
while($row = mysqli_fetch_assoc($aresult)){
echo "['".$row["Week"]."', ".$row["Omean"]."],";
}
?>
]);
var view = new google.visualization.DataView(data);
view.setColumns([0, 1, {
calc: 'stringify',
sourceColumn: 1,
type: 'string',
role: 'annotation'
}]);
var options = {
title: 'Chart1',
'width':1700,
'height':600,
hAxis : {textStyle : {fontSize: 8}},
'tooltip' : { trigger: 'none'},
enableInteractivity: false,
legend: { position: 'bottom' }
};
var chart = new
google.visualization.LineChart(document.getElementById('curvechart'));
chart.draw(data, options);
chart.draw(view, options); // <-- use data view
}
</script>
</head>
<body>
<div id="curvechart" style="width: 900px; height: 400px"></div>
</body>
</html>
图表1
load('current',{'packages':['corechart']});
google.charts.setOnLoadCallback(drawChart);
函数绘图图(){
var data=new google.visualization.DataTable();
var data=google.visualization.arrayToDataTable([
[‘周’、‘欧米安’],
“两个按钮都不起作用”这意味着什么?单击时没有触发?错误的结果?没有结果?是的,当我单击按钮时,没有结果显示。chart.php页面自行工作。您不能通过chart1.php从ajax加载图表。如果您想加载图表,您需要一个json返回值并将其加载到chart.php而不是chart1.phpok中,谢谢,NigHamza。