Question

我有以下坐标数组：

var coords = [35.77204705542798, -5.815865197320899, 35.77205120747819, -5.815754188240848, 35.77197468036722, -5.815810097181759, 35.77201185726312, -5.816182008817898, 35.77188028308802, -5.816782866625928, 35.77144809183601, -5.817054836919457, 35.77130127311978, -5.817231221149015, 35.77121654704168, -5.817323279458099, 35.77197602552531, -5.818491135420929, 35.77211527443405, -5.818831898394636, 35.77220613582161, -5.819054426882189, 35.77225761824354, -5.819182167400329, 35.77557809840525, -5.819685065789988, 35.77522352494348, -5.821346612263469, 35.77815743782872, -5.822735160242799, 35.77863231735067, -5.822959942846667, 35.77915215012052, -5.823206036666789, 35.77917987363854, -5.82321907562955, 35.77913908191616, -5.825751400638378, 35.77947725976961, -5.826044519616627, 35.77923063670355, -5.82893344672563, 35.77920722271806, -5.829544463218058, 35.77924325929096, -5.82965650991142, 35.77913169320556, -5.830906964998152, 35.77917349450947, -5.830173084585434, 35.78563643141488, -5.829373128632887, 35.7859055726137, -5.829317961099034, 35.78659089504106, -5.829249272155759, 35.78704770052305, -5.829163948962961, 35.78696081760869, -5.829114610331397, 35.78687737614365, -5.829067226012621, 35.78679073445069, -5.829018024365507, 35.78661289235281, -5.828929232334034, 35.78652027010703, -5.828884857172518, 35.78643576729606, -5.828844387274908, 35.78635159916791, -5.828804167105649, 35.7862672257822, -5.828763857888811, 35.7872100444153, -5.866787159467321]

基本上，订单类似于以下([lat,lng,lat,lng........])

var waypts = [];

//should execute for loop here, get lat and lng from the array
stop = new google.maps.LatLng(lat, lng)
waypts.push({
    location: stop,
    stopover: true
});
///

我已尝试过以下内容，但它似乎只运行一次

for(var i=0; i<=coords.length; i++){
//get only even so I can have something like coords.slice(0,2), coords.slice(2,4)
if ((i % 2) == 0){
coords.slice(i+0,i+2)
 }
}

将在后台运行的最终结果应类似于：

stop = new google.maps.LatLng(35.77205120747819, -5.815754188240848)
waypts.push({
    location: stop,
    stopover: true
});

stop = new google.maps.LatLng(35.77197468036722, -5.815810097181759)
waypts.push({
    location: stop,
    stopover: true
});

Answer 1

您可以遍历每个对：

mergeProps

Answer 2

无需进行模块化数学运算 - 只需使用https和i（这将增加++i并且循环将正常处理）

Answer 3

看看这个工作示例，我减少了坐标的数量，使其更具可读性。

＆＃13;

var coords = [35.77204705542798, -5.815865197320899, 35.77205120747819, -5.815754188240848, 35.77197468036722, -5.815810097181759];

for (var i = 0; i <= coords.length - 2; i++) {
  if ((i % 2) == 0) {
    console.log("Lat:" + coords[i]);
    console.log("Lon:" + coords[i + 1]);
    /*stop = new google.maps.LatLng(coords[i], coords[i + 1])
    waypts.push({
      location: stop,
      stopover: true
    });*/
  }
}

＆＃13;

Answer 4

您可以使用while和shift它非常干净。

while(coords.length) stop = new google.maps.LatLng(coords.shift(), coords.shift());

Answer 5

您可以将coords数组转换为一个对象数组，每个对象包含一个lat和long变量。我为每个纬度/经度使用不同的值，使其不会那么长。

var coords = [{lat:1,long:2},{lat:3,long:4},{lat:5,long:6},{lat:7,long:8}];
for(var i=0; i<=coords.length; i++) {
    // perform rest of operation here
}

JavaScript - 以特定方式组织数组

5 个答案: