67

Is it possible to make an ajax request inside another ajax request? because I need some data from first ajax request to make the next ajax request.

First I'm using Google Maps API to get LAT & LNG, after that I use that LAT & LNG to request Instagram API (search based location).

Once again, is this possible, and if so how?

$('input#search').click(function(e) {
    e.preventDefault();
    var source = $('select[name=state] option:selected').text()+' '+$('select[name=city] option:selected').text()+' '+$('select[name=area] option:selected').text();
    var source = source.replace(/ /g, '+');
    if(working == false) {
        working = true;
        $(this).replaceWith('<span id="big_loading"></span>');
        $.ajax({
            type:'POST',
            url:'/killtime_local/ajax/location/maps.json',
            dataType:'json',
            cache: false,
            data:'via=ajax&address='+source,
            success:function(results) {
            // this is where i get the latlng
            }
        });
    } else {
        alert('please, be patient!');
    }
});
4

4 回答 4

96

这是一个例子:

$.ajax({
    type: "post",
    url: "ajax/example.php",
    data: 'page=' + btn_page,
    success: function (data) {
        var a = data; // This line shows error.
        $.ajax({
            type: "post",
            url: "example.php",
            data: 'page=' + a,
            success: function (data) {
   
            }
        });
    }
});
于 2012-04-10T13:19:14.300 回答
23

从“完成”调用第二个 ajax

这是示例

   var dt='';
   $.ajax({
    type: "post",
    url: "ajax/example.php",
    data: 'page='+btn_page,
    success: function(data){
        dt=data;
        /*Do something*/
    },
    complete:function(){
        $.ajax({
           var a=dt; // This line shows error.
           type: "post",
           url: "example.php",
           data: 'page='+a,
           success: function(data){
              /*do some thing in second function*/
           },
       });
    }
});
于 2015-07-04T04:34:47.627 回答
4

这只是一个例子。您可能希望根据您的要求对其进行自定义。

 $.ajax({
      url: 'ajax/test1.html',
      success: function(data1) {
        alert('Request 1 was performed.');
        $.ajax({
            type: 'POST',
            url: url,
            data: data1, //pass data1 to second request
            success: successHandler, // handler if second request succeeds 
            dataType: dataType
        });
    }
});

有关更多详细信息:请参阅

于 2012-04-10T13:24:13.497 回答
0
$.ajax({
    url: "<?php echo site_url('upToWeb/ajax_edit/')?>/" + id,
    type: "GET",
    dataType: "JSON",
    success: function (data) {
        if (data.web == 0) {
            if (confirm('Data product upToWeb ?')) {
                $.ajax({
                    url: "<?php echo site_url('upToWeb/set_web/')?>/" + data.id_item,
                    type: "post",
                    dataType: "json",
                    data: {web: 1},
                    success: function (respons) {
                        location.href = location.pathname;
                    },
                    error: function (xhr, ajaxOptions, thrownError) { // Ketika terjadi error
                        alert(xhr.responseText); // munculkan alert
                    }
                });
            }
        }
        else {
            if (confirm('Data product DownFromWeb ?')) {
                $.ajax({
                    url: "<?php echo site_url('upToWeb/set_web/')?>/" + data.id_item,
                    type: "post",
                    dataType: "json",
                    data: {web: 0},
                    success: function (respons) {
                        location.href = location.pathname;
                    },
                    error: function (xhr, ajaxOptions, thrownError) { // Ketika terjadi error
                        alert(xhr.responseText); // munculkan alert
                    }
                });
            }
        }
    },

    error: function (jqXHR, textStatus, errorThrown) {
        alert('Error get data from ajax');
    }

});
于 2017-07-30T02:58:47.170 回答