jQuery ajax将变量发布到php

Question

my ajax.php script 我的ajax.php脚本

<?php
$lat =$_POST['lat'];
$lon = $_POST['lon'];
echo $lat;
echo $lon;
?>

And my new.html page 还有我的new.html页面

<!DOCTYPE html>
<html><head><script src="jquery-1.12.0.min.js"></script>
</head>
<body>

<p>Click the button to get your coordinates.</p>
<input type="text" name="metin"/><br/>
<button onclick="getLocation()">Try It</button>

<p id="demo"></p>


<p id="demo1"></p>
<script>

var x = document.getElementById("demo");
var y = document.getElementById("demo1");

function getLocation() {
    if (navigator.geolocation) {
        navigator.geolocation.getCurrentPosition(showPosition);
    } else { 
        x.innerHTML = "Geolocation is not supported by this browser.";
    }
}
function showPosition(position) {
    x.innerHTML = "Latitude: " + position.coords.latitude + "<br>Longitude: " + position.coords.longitude;
    var lon= position.coords.longitude;
    var lat= position.coords.longitude;
    y.innerHTML=lon;
    x.innerHTML=lat;
    $.ajax({
            type: 'POST',
            url: 'ajax.php',
            data:  { lat:lat, lon:lon}
             success: function(result) {
                $('#sonuc').html(result);
            },
            error: function() {
                alert('Some error found. Please try again!');
            }
        });
    }
}

</script>

<p id="sonuc"></p>
</body>
</html>

Firstly i know there is unused codes. 首先，我知道有未使用的代码。 But İ'm gonna use this blocks. 但是İ将使用此块。 İt's not give me any response. 不给我任何答复。 İt's have to work successfuly. 它必须成功地工作。 But it's not give me lat and lon varibles in my php page. 但这并没有在我的php页面中给我lat和lon变量。 İf there is another way to send variables to php page i can try. 还有一种将变量发送到php页面的方法，我可以尝试。

Answer 1

No functional issues in code. 代码中没有功能性问题。 But there is some syntax errors that blocks the ajax request. 但是有一些语法错误会阻止ajax请求。

Avoid the extra closing brace below the function showPosition(). 避免在函数showPosition（）下使用多余的右括号。
Add comma after the data section in ajax request. 在ajax请求中的数据部分之后添加逗号。

Corrected code is given below. 更正的代码如下。

 <script> var x = document.getElementById("demo"); var y = document.getElementById("demo1"); function getLocation() { if (navigator.geolocation) { navigator.geolocation.getCurrentPosition(showPosition); } else { x.innerHTML = "Geolocation is not supported by this browser."; } } function showPosition(position) { x.innerHTML = "Latitude: " + position.coords.latitude + "<br>Longitude: " + position.coords.longitude; var lon= position.coords.longitude; var lat= position.coords.longitude; y.innerHTML=lon; x.innerHTML=lat; $.ajax({ type: 'POST', url: 'ajax.php', data: { lat:lat, lon:lon}, success: function(result) { $('#sonuc').html(result); }, error: function() { alert('Some error found. Please try again!'); } }); } </script>

jQuery ajax将变量发布到php

问题描述

1 个解决方案

解决方案1
2 已采纳 2016-04-03 19:51:07

jQuery ajax将变量发布到php

问题描述

1 个解决方案

解决方案1 2 已采纳 2016-04-03 19:51:07

解决方案1
2 已采纳 2016-04-03 19:51:07