Question

如何将Javascript变量传递给Ajax，然后将其发布到php。 Javacript代码和PHP代码可以正常工作，但问题是将Javascript变量传递给Ajax以发布到PHP。我的代码在

下

<!-- User Location --> 
<script>
window.onload = function() {
    var startPos;

    var geoSuccess = function(position) {
        startPos = position;
        document.getElementById('startLat').innerHTML = startPos.coords.latitude;
        document.getElementById('startLon').innerHTML = startPos.coords.longitude; 

        $.ajax({
            type: 'POST', 
            url: 'MyDashboard.php', 
            data: 'latitude='+latitude+'&longitude='+longitude, 
            success: function(msg) { 
                if (msg) { 
                    $("#location").html(msg); 
                } else { 
                    $("#location").html('Not Available'); 
                } 
            } 
        });
    };

    navigator.geolocation.getCurrentPosition(geoSuccess);
};
</script>

Answer 1

不确定您要问什么但只是

data:`latitude=${startPos.coords.latitude}&longitude=${startPos.coords.longitude}`

Answer 2

您应该将数据作为USE my_target_db; SET @origin_db='my_origin_db'; SET @origin_table = CONCAT(@origin_db,'.','tablename'); SET @query = CONCAT('INSERT INTO target_table SELECT * FROM ', @origin_table); PREPARE stmt FROM @query; EXECUTE stmt; DEALLOCATE PREPARE stmt;之类的对象进行传递

data: {latitude: latitude, longitude:longitude }

并输入<script> window.onload = function() { var startPos; var geoSuccess = function(position) { startPos = position; document.getElementById('startLat').innerHTML = startPos.coords.latitude; document.getElementById('startLon').innerHTML = startPos.coords.longitude; $.ajax({ type: 'POST', url: 'MyDashboard.php', data: {latitude: startPos.coords.latitude, longitude: startPos.coords.longitude }, success: function(msg) { if (msg) { $("#location").html(msg); } else { $("#location").html('Not Available'); } } }); }; navigator.geolocation.getCurrentPosition(geoSuccess); }; </script>和$_POST['latitude']之类的php代码

Answer 3

问题是latitude和longitude变量不存在。.

尝试以下代码：

<!-- User Location -->
<script>
  window.onload = function() {
    var geoSuccess = function(position) {
      document.getElementById('startLat').innerHTML = position.coords.latitude;
      document.getElementById('startLon').innerHTML = position.coords.longitude;

      $.ajax({
        type: 'POST',
        url: 'MyDashboard.php',
        data: 'latitude=' + position.coords.latitude+ '&longitude=' + position.coords.latitude,
        success: function(msg) {

          if (msg) {
            $("#location").html(msg);
          } else {
            $("#location").html('Not Available');
          }
        }

      });
    };
    navigator.geolocation.getCurrentPosition(geoSuccess);
  };
</script>

并通过调用这些变量$_POST['latitude']和$_POST['longitude']

在PHP端检索变量

Answer 4

您正在传递未定义的变量。请尝试这个：

$.ajax({
    type: 'POST',
    url: 'MyDashboard.php',
    data: {
      latitude: startPos.coords.latitude,
      longitude: startPos.coords.longitude
    },
    success: function(msg) {

    // your logic

});

Answer 5

您将错误的数据发送到AJAX POST方法

数据：{'纬度'：纬度，'经度'：经度}

发送AJAX的位置

数据：'latitude ='+纬度+'＆longitude ='+经度，

或者您可以在URL上添加它们

url：'MyDashboard.php？latitude ='+纬度+'＆longitude ='+经度

这也将起作用，但是$ _GET需要在php中使用$ _POST

将Javascript GetElementById变量传递给Ajax

5 个答案: