Question

我在php中有一个数组（$ coords），其中包含一些坐标。

print_r($coords)

给出

Array ( [0] => 80.23338342111089 ,26.52748985468579 [1] => 80.21915925983853 ,26.51065783417543 [2] => 80.23036859334592 ,26.49785585262899 [3] => 80.24667293123426 ,26.51075259323626 [4] => 80.23338342111089 ,26.52748985468579 )

然后我使用了$string = implode(",",$coords); print_r($string);

它给了我

80.23338342111089 ,26.52748985468579,80.21915925983853 ,26.51065783417543,80.23036859334592 ,26.49785585262899,80.24667293123426 ,26.51075259323626,80.23338342111089 ,26.52748985468579

我将此传递给Javascript，就像这样 -

    var Coordinates = [];
    Coordinates.push(<?php echo ($string); ?>);

现在我想将其字符串化，所以我使用了 -

var JSON_Coordinates = JSON.stringify(Coordinates);

现在JSON_Coordinates的值是 -

[80.23338342111089 ,26.52748985468579,80.21915925983853 ,26.51065783417543,80.23036859334592 ,26.49785585262899,80.24667293123426 ,26.51075259323626,80.23338342111089 ,26.52748985468579]

但它应该是 -

  [[80.23338342111089 ,26.52748985468579],[80.21915925983853 ,26.51065783417543],[80.23036859334592 ,26.49785585262899],[80.24667293123426 ,26.51075259323626],[80.23338342111089 ,26.52748985468579]]

有人请告诉我我哪里错了？

感谢。

Answer 1

使用array_chunk将数组拆分为

$arr = [80.23338342111089 ,26.52748985468579,80.21915925983853 ,26.51065783417543,80.23036859334592 ,26.49785585262899,80.24667293123426 ,26.51075259323626,80.23338342111089 ,26.52748985468579];
?>
var Coordinates = <?= json_encode(array_chunk($arr,2)) ?>;
// var Coordinates = [[80.233383421111,26.527489854686],[80.219159259839,26.510657834175],[80.230368593346,26.497855852629],[80.246672931234,26.510752593236],[80.233383421111,26.527489854686]];

demo

Answer 2

需要了解如何在PHP代码中建立数组。它们是成对出现还是全部放在一个大阵列中？

非常简单。只是与您要转换为JSON的数组的结构有关。

$foo = array(1, 2, 3, 4);
// [1,2,3,4]

$bar = array(array(1, 2), array(3, 4));
// [[1,2],[3,4]]

我清理数据服务器端，所以你只需要对其进行JSON编码，然后通过该客户端。

Answer 3

您应该像这样编写JS逻辑。此时，您在代码中什么都不做，将创建所需的2d数组。

你应该循环遍历整个数组并将计数器递增2以一次获得2个元素并创建一个新的临时数组，该数组将推入原始的arrCoordinates。

协调供应为数组

＆＃13;

var coordinates = [80.23338342111089 ,26.52748985468579,80.21915925983853 ,26.51065783417543,80.23036859334592 ,26.49785585262899,80.24667293123426 ,26.51075259323626,80.23338342111089 ,26.52748985468579];

var arrCoordinates =[];
for(var index=0; index< coordinates.length; index+=2){
    arrCoordinates.push([coordinates[index], coordinates[index+1]]);
}

console.log(arrCoordinates);

＆＃13;

将供应协调为字符串

＆＃13;

var coordinates =  "80.23338342111089 ,26.52748985468579,80.21915925983853 ,26.51065783417543,80.23036859334592 ,26.49785585262899,80.24667293123426 ,26.51075259323626,80.23338342111089 ,26.52748985468579";

var arrCoordinates =[];
var inputCoordinates = coordinates.split(",");
for(var index=0; index< inputCoordinates.length; index+=2){
    arrCoordinates.push([inputCoordinates[index].trim(), inputCoordinates[index+1].trim()]);
}

console.log(arrCoordinates);

＆＃13;

Answer 4

获得坐标数组后，可以循环遍历它以构建所需的输出。

另一种方法是使用for-loop。

＆＃13;

var coordinates = [80.23338342111089 ,26.52748985468579,80.21915925983853 ,26.51065783417543,80.23036859334592 ,26.49785585262899,80.24667293123426 ,26.51075259323626,80.23338342111089 ,26.52748985468579];

var newCoordinates = [];
for (var i = 0; i < coordinates.length; i++) {
  if (i % 2 !== 0) newCoordinates.push([coordinates[i - 1], coordinates[i]]);
}

console.log(newCoordinates);

＆＃13;

.as-console-wrapper { max-height: 100% !important; top: 0; }

＆＃13;

根据您的初始数组，您可以循环，拆分然后构建嵌套数组：

初始输入数据： ["c1, c2", "c1, c2", ...]

这样，您就不需要调用函数stringify。

＆＃13;

var array = ["80.23338342111089 ,26.52748985468579", "80.21915925983853 ,26.51065783417543","80.23036859334592 ,26.49785585262899", "80.24667293123426 ,26.51075259323626","80.23338342111089 ,26.52748985468579" ];

var result = [];
for(var i = 0; i < array.length; i++) {
  var split = array[i].split(',');
  var lt = split[0].trim();
  var ln = split[1].trim();
  result.push([+lt.trim(), +ln.trim()])
}

console.log(result);

＆＃13;

.as-console-wrapper { max-height: 100% !important; top: 0; }

＆＃13;

php变量上的JSON.stringify

4 个答案: