无法使用AJAX将变量发布到PHP
[英]Can't post variable to PHP using AJAX
问题描述
I am trying to send a variable to PHP using JavaScript but I don't get any response whatsoever. 
我正在尝试使用JavaScript将变量发送到PHP，但是却没有任何响应。 Any ideas? 有任何想法吗？
PHP/HTML: PHP / HTML：
<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>
<script src="dashmenu.js"></script>
</head>
<body>
<?php
$selection = $_POST['selection'];
if($selection == 'profile'){
?>
<p> it works
<?php
}
?>
<button id="button"> profile </button>
</body>
JS file (dashmenu.js): JS档案（dashmenu.js）：
$('#button').click(function() {
var menuSelection = "profile";
$.ajax({
type: 'POST',
url: 'dashboard.php',
data: {selection: menuSelection},
success: function(response) {
alert('success');
}
});
});
2 个解决方案
解决方案1
0 2016-05-19 20:09:31
In your html file (let's say index.html) you could have: 在您的html文件中（假设为index.html），您可以：
$('#button').click(function() {
var menuSelection = "profile";
$.ajax({
type: 'POST',
dataType: "html",
url: 'dashboard.php',
data: {selection: menuSelection},
success: function(response) {
alert(response);
},error: function(jqXHR, textStatus, errorThrown){
alert('Error: ' + errorThrown);
}
});
});
In your dashboard.php, you should ONLY have code that processes requests, and returns (echoes) the desired output: 在dashboard.php中，您仅应具有处理请求并返回（回显）所需输出的代码：
<?php
$selection = $_POST['selection'];
if($selection == 'profile'){
echo "It works";
}else{
echo "It failed";
}
?>
解决方案2
0 2016-05-20 05:58:26
$('#button').click(function() { var menuSelection = "profile"; $（'＃button'）。click（function（）{var menuSelection =“ profile”;
$.ajax({
type: 'POST',
url: 'dashboard.php',
data: {selection: menuSelection},
success: function(response) {
alert('success');
}
});
}); }）;
Run this in your console in your browser. 在浏览器的控制台中运行此命令。 Right-click > Console tab. 右键单击>控制台选项卡。 If you want to check whether this function has successfully bind to the button. 如果要检查此功能是否已成功绑定到按钮。 If your browser returns you 'success' alert, meaning you include it wrongly I guess. 如果您的浏览器返回“成功”警报，则意味着您错误地包含了它。
If nothing happen when you click, please include .fail() in your $.ajax() 如果单击时没有任何反应，请在$ .ajax（）中包括.fail（）
Which will look like this: 看起来像这样：
$.ajax({
type: 'POST',
url: 'dashboard.php',
data: {
selection: menuSelection
},
success: function(response) {
alert(response);
},
error: function(jqXHR, textStatus, errorThrown){
}
});
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