javascript - 删除所有其他逗号并用方括号括起来

Question

所以使用 jquery，我得到了一串这样的坐标:

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空格和括号坐标集替换所有其他逗号。

想法？

Answer 1

您可以使用正则表达式。

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

Regex Explanation and Live Demo

1. [-\d.]:字符类，- 将匹配文字 - 连字符，\d将匹配单个数字，. 将匹配 . 字面意思。在类中提到时，顺序无关紧要。
2. +:匹配前面匹配项的一次或多次出现
3. (...):捕获组。大括号内的匹配被捕获并在 $1, $2, ...
中返回
4. ,?:不匹配每隔一个逗号
5. 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, '[$1, $2] ').trim();
document.getElementById('output').innerHTML = JSON.stringify(result, 0, 2);

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