删除所有其他逗号并用方括号括起来

Question

所以使用 jquery 我得到了一串坐标，如下所示：

38.313072,-89.863845,38.312675,-89.863586,38.310405,-89.862091,38.310405,-89.862091,38.309913,-89.861976,38.309768,-89.861976,38.309768,-89.861976

我需要它们看起来像这样：

[38.313072, -89.863845] [38.312675, -89.863586] [38.310405, -89.862091] [38.310405,-89.862091] [38.309913, -89.861976] [38.309768, -89.861976] [38.309768, -89.861976] [38.30965, -89.861991]

所以我需要弄清楚如何用 space 和括号坐标集替换所有其他逗号。

想法？

Answer 1

不漂亮，但它有效：

var coordinateString = "38.313072,-89.863845,38.312675,-89.863586,38.310405,-89.862091,38.310405,-89.862091,38.309913,-89.861976,38.309768,-89.861976,38.309768,-89.861976";

var coordinateArray = coordinateString.split(",");
var coordinateResult = "";
var i = 0;
while(i<coordinateArray.length){
  coordinateResult +="[" + coordinateArray[i] + ", " + coordinateArray[i+1] + "]";
  i += 2;
}

Answer 2

您可以使用正则表达式。

([-\d.]+),([-\d.]+),?

Regex Explanation and Live Demo

[-\d.]：字符class，-将匹配文字-连字符，\d将匹配单个数字，.将按字面匹配 .。当在 class 中提到时，顺序无关紧要。
+：匹配一个或多个先前匹配项
(...)：捕获组。捕获大括号内的匹配项并在 </code>、<code>、...
,?：不匹配每隔一个逗号
g：全局匹配。匹配所有可能出现的情况。

jsFiddle Demo

var str = '38.313072,-89.863845,38.312675,-89.863586,38.310405,-89.862091,38.310405,-89.862091,38.309913,-89.861976,38.309768,-89.861976,38.309768,-89.861976';

var result = str.replace(/([-\d.]+),([-\d.]+),?/g, '[, ] ').trim();
document.getElementById('output').innerHTML = JSON.stringify(result, 0, 2);

<pre id="output"></pre>

Answer 3

这应该可以完成工作。

function toPairs(src){
  // Split the string into values.
  var values = src.split(',');
  
  // Group these values 2 by 2.
  var pairs = [];
  for(var i = 0; i < values.length; i += 2){
    pairs.push("[" + values[i] + ", " + values[i + 1] + "]");
  }
  
  // Join with a whitespace.
  return pairs.join(" ");
}

document.body.innerHTML = toPairs("38.313072,-89.863845,38.312675,-89.863586,38.310405,-89.862091,38.310405,-89.862091,38.309913,-89.861976,38.309768,-89.861976,38.309768,-89.861976");

Answer 4

尝试利用 while 循环，Array.prototype.splice()，以数组数组的形式返回结果

var str = "38.313072,-89.863845,38.312675,-89.863586,38.310405,-89.862091,38.310405,-89.862091,38.309913,-89.861976,38.309768,-89.861976,38.309768,-89.861976"
, res = []
, arr = str.split(",");
while (arr.length) res.push(arr.splice(0, 2))

console.log(res, JSON.stringify(res, null, 2))

Answer 5

使用正则表达式！非常适合这个！

var str = "38.313072,-89.863845,38.312675,-89.863586,38.310405,-89.862091,38.310405,-89.862091,38.309913,-89.861976,38.309768,-89.861976,38.309768,-89.861976,38.30965, -89.861991";

var expected = "[38.313072, -89.863845] [38.312675, -89.863586] [38.310405, -89.862091] [38.310405,-89.862091] [38.309913, -89.861976] [38.309768, -89.861976] [38.309768, -89.861976] [38.30965, -89.861991]"


var computed = str.replace(/(([0-9\-. ]*)(,)([0-9\-. ]*)),?/g, '[  ] ');

document.write('computed<br>')
document.write(str.replace(/(([0-9\-. ]*)(,)([0-9\-. ]*)),?/g, '[  ] '))



document.write( "<hr>expected<br>" + expected  )

删除所有其他逗号并用方括号括起来

Remove every other comma and surround with brackets

javascript

gis

jquery

coordinates