The following code should add 100 to an existing number in a mysql table if the button gets clicked. If I click the button nothing happens, but if I reload the page the function adds 100 to the number. What is wrong with my code?
如果单击该按钮，则以下代码应将mysql表中的现有数字添加100。如果我单击按钮没有任何反应，但如果我重新加载页面，该功能会为该数字添加100。我的代码出了什么问题？
<?php
define('DBHOST', 'localhost');
define('DBUSER', 'root');
define('DBPASS', '123');
define('DBNAME', 'dbtest');
$conn = mysql_connect(DBHOST,DBUSER,DBPASS);
$dbcon = mysql_select_db(DBNAME);
?>
<!DOCTYPE html>
<html lang="en">
<head>
<title>Test</title>
<meta charset="utf-8">
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.1/jquery.min.js"></script>
<link rel="stylesheet" href="https://cdnjs.cloudflare.com/ajax/libs/materialize/0.97.8/css/materialize.min.css">
</head>
<body>
<a id="button" class="waves-effect btn deep-orange darken-1">Button 1</a>
</body>
<script>
$("#button").click(function(){
<?php
mysql_query("UPDATE users SET test = (test + 100) WHERE userId=1");
?>
});
</script>
</html>
2 个解决方案
#1
1
You cant call PHP code from a jQuery function like that. All the php runs when the page loads and thats it. You can however use jQuery and Ajax to send a message to a php script that processes that message then returns a response. The script can even be in the same actual file like you have (or in a different file altogether) something like this would do:
你不能从像这样的jQuery函数调用PHP代码。当页面加载时，所有的PHP都会运行。但是，您可以使用jQuery和Ajax将消息发送到处理该消息然后返回响应的php脚本。该脚本甚至可以像你一样在同一个实际文件中（或者在一个不同的文件中），就像这样：
<?php
define('DBHOST', 'localhost');
define('DBUSER', 'root');
define('DBPASS', '123');
define('DBNAME', 'dbtest');
$conn = mysql_connect(DBHOST,DBUSER,DBPASS);
$dbcon = mysql_select_db(DBNAME);
if(isset($_POST['updateTest']){
$val = $_POST['test'];
$id + $_POST['userId'];
// validate inputs and such....
mysql_query("UPDATE users SET test = (test + 100) WHERE userId=1");
// send success or error response...
echo json_encode(['success'=>true]);
exit;
}
?>
<!DOCTYPE html>
<html lang="en">
<head>
<title>Test</title>
<meta charset="utf-8">
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.1/jquery.min.js"></script>
<link rel="stylesheet" href="https://cdnjs.cloudflare.com/ajax/libs/materialize/0.97.8/css/materialize.min.css">
</head>
<body>
<a id="button" class="waves-effect btn deep-orange darken-1">Button 1</a>
</body>
<script>
$("#button").click(function(){
var count = 100;
var userId = 1;
var dataObject= {updateTest: true, test: 100, userId: 1};
$.ajax({
type: "POST",
// url: "page.php", // add this line to send to some page other than the this one
data: dataObject,
success: function(response) {
if(response.success){
alert('test worked');
}
else{
alert('there was an error')
}
},
error: function(xhr, status, error) {
console.log(xhr);
}
});
});
</script>
</html>
#2
1
As mentioned by the previous poster PHP is server side and Javascript client side so what is actually happening is the following.
正如之前的海报所提到的，PHP是服务器端和Javascript客户端，所以实际发生的是以下内容。
When the page is returned back to the user your piece of javascript just looks like the below..
当页面返回给用户时，你的javascript就像下面的那样......
Your MySQL statement here has executed already it can not interact with client side code in this way
你的MySQL语句已经执行，它已经无法以这种方式与客户端代码交互
<script>
$("#button").click(function(){
// nothing here.. But your MYSQL statement has executed anyway
});
</script>
#1
1
You cant call PHP code from a jQuery function like that. All the php runs when the page loads and thats it. You can however use jQuery and Ajax to send a message to a php script that processes that message then returns a response. The script can even be in the same actual file like you have (or in a different file altogether) something like this would do:
你不能从像这样的jQuery函数调用PHP代码。当页面加载时，所有的PHP都会运行。但是，您可以使用jQuery和Ajax将消息发送到处理该消息然后返回响应的php脚本。该脚本甚至可以像你一样在同一个实际文件中（或者在一个不同的文件中），就像这样：
<?php
define('DBHOST', 'localhost');
define('DBUSER', 'root');
define('DBPASS', '123');
define('DBNAME', 'dbtest');
$conn = mysql_connect(DBHOST,DBUSER,DBPASS);
$dbcon = mysql_select_db(DBNAME);
if(isset($_POST['updateTest']){
$val = $_POST['test'];
$id + $_POST['userId'];
// validate inputs and such....
mysql_query("UPDATE users SET test = (test + 100) WHERE userId=1");
// send success or error response...
echo json_encode(['success'=>true]);
exit;
}
?>
<!DOCTYPE html>
<html lang="en">
<head>
<title>Test</title>
<meta charset="utf-8">
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.1/jquery.min.js"></script>
<link rel="stylesheet" href="https://cdnjs.cloudflare.com/ajax/libs/materialize/0.97.8/css/materialize.min.css">
</head>
<body>
<a id="button" class="waves-effect btn deep-orange darken-1">Button 1</a>
</body>
<script>
$("#button").click(function(){
var count = 100;
var userId = 1;
var dataObject= {updateTest: true, test: 100, userId: 1};
$.ajax({
type: "POST",
// url: "page.php", // add this line to send to some page other than the this one
data: dataObject,
success: function(response) {
if(response.success){
alert('test worked');
}
else{
alert('there was an error')
}
},
error: function(xhr, status, error) {
console.log(xhr);
}
});
});
</script>
</html>
#2
1
As mentioned by the previous poster PHP is server side and Javascript client side so what is actually happening is the following.
正如之前的海报所提到的，PHP是服务器端和Javascript客户端，所以实际发生的是以下内容。
When the page is returned back to the user your piece of javascript just looks like the below..
当页面返回给用户时，你的javascript就像下面的那样......
Your MySQL statement here has executed already it can not interact with client side code in this way
你的MySQL语句已经执行，它已经无法以这种方式与客户端代码交互
<script>
$("#button").click(function(){
// nothing here.. But your MYSQL statement has executed anyway
});
</script>