小编Der*_*erk的帖子

将参数传递给回调函数

我的代码

//执行ajax请求并获取JSON响应

for (var i = 0; i < data.results.length; i++) {  
    result = data.results[i];
    // do stuff and create google maps marker    
    marker = new google.maps.Marker({  
        position: new google.maps.LatLng(result.lat, result.lng),   
        map: map,  
        id: result.id  
    });  
    google.maps.event.addListener(marker, 'click', function() {  
        createWindow(marker.id); //<==== this doesn't work because marker always points to the last results when this function is called
    });  

}
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怎么解决这个?

javascript google-maps dom-events

10
推荐指数
1
解决办法
1万
查看次数

调试jQuery AJAX响应:我做错了什么?

$.ajax({
    type: 'POST',
    url: 'place/add',
    data: {
        lat: lat,
        lng: lng,
        name: name,
        address: address,
        phone: phone,
        review: review,
        category: category
    },
    success: function(data) {
    alert(data);
    alert(data.id);
    // ......
});
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第一个警报给出:{"id":"2","success":true}但第二个:undefined

javascript debugging ajax jquery json

2
推荐指数
1
解决办法
928
查看次数

标签 统计

javascript ×2

ajax ×1

debugging ×1

dom-events ×1

google-maps ×1

jquery ×1

json ×1