Question

所以这个问题以前曾被问过，但是我的情况有点不同，因为我已经尝试了该网站中建议的所有解决方案，所以我有一个多维度数组：

var array = [
{
    "dup":{ "lat":36.69507771893455,"lng":2.828052824475776 },
    "original":{ "lat":36.695076,"lng":2.8280580000000555}
},
{
    "dup":{ "lat":36.69507687630523,"lng":2.8280525056927672},
    "original":{ "lat":36.695076,"lng":2.8280580000000555}
},
{
    "dup":{ "lat":36.695076,"lng":2.8280523980529324},
    "original":{ "lat":36.695076,"lng":2.8280580000000555}
},
{
    "dup":{ "lat":36.69507512369477,"lng":2.8280525056927672},
    "original":{ "lat":36.695076,"lng":2.8280580000000555}
},
{
    "dup":{ "lat":36.69507428106545,"lng":2.828052824475776},
    "original":{ "lat":36.695076,"lng":2.8280580000000555}
},
{
    "dup":{ "lat":36.695073504493834,"lng":2.828053342151293},
    "original":{ "lat":36.695076,"lng":2.8280580000000555}
},
...
]

还有另一个小数组：

var datax = {"lat":36.69507428106545,"lng":2.828052824475776}

我想要做的就是找到与此“ dup”在同一行中的“原始”。到目前为止，我尝试过查找，过滤和grep，但这都不能解决我的问题。

我尝试过的代码：

var Cent = $.grep(array, function(e){ return e.dup== JSON.stringify(datax); });
contentString += " and "+Cent +"is the closer";

Answer 1

您仍然可以使用Array.filter()来实现：

var array = [{"dup":{"lat":36.69507771893455,"lng":2.828052824475776},"original":{"lat":36.695076,"lng":2.8280580000000555}},{"dup":{"lat":36.69507687630523,"lng":2.8280525056927672},"original":{"lat":36.695076,"lng":2.8280580000000555}},{"dup":{"lat":36.695076,"lng":2.8280523980529324},"original":{"lat":36.69507428106545,"lng":2.828052824475776}},{"dup":{"lat":36.69507512369477,"lng":2.8280525056927672},"original":{"lat":36.695076,"lng":2.8280580000000555}},{"dup":{"lat":36.69507428106545,"lng":2.828052824475776},"original":{"lat":36.695076,"lng":2.8280580000000555}},{"dup":{"lat":36.695073504493834,"lng":2.828053342151293},"original":{"lat":36.695076,"lng":2.8280580000000555}}];
var datax = {"lat":36.69507428106545,"lng":2.828052824475776}

var res = array.filter(item => item.original.lat === datax.lat && item.original.lng === datax.lng);
console.log(res);

在JavaScript数组中查找对象

1 个答案: