正则表达式匹配第二个下划线后的所有内容

Question

我有这个数组

var collUnlocked = ['GRAVEL_1', 'GRAVEL_3', 'GRAVEL_2', 'GRAVEL_5', 'GRAVEL_4', 'GRAVEL_10', 'GRAVEL_-1', 'LOG:3_1', 'LOG:3_-1', 'LOG:3_3', 'LOG:3_4', 'LOG_2_6', 'LOG_2_2', 'LOG_2_3', 'LOG_2_-1']

我想获取姓名（例如 LOG_2）和最后一个数字（例如 -1）

示例： ['LOG:3_4'、'GRAVEL_-1'、'LOG_2_-1']

变成 -> [['LOG:3','4'], ['GRAVEL', '-1'], ['LOG_2', '-1']]

编辑：我正在使用此代码获取姓名，但我无法获取最后一个号码

collName = x.match('(.*\_)')[0];
collName = collName.slice(0, -1);

完整代码 rn:

function testingAll() {
  var collUnlocked = ['GRAVEL_1', 'GRAVEL_3', 'GRAVEL_2', 'GRAVEL_5', 'GRAVEL_4', 'GRAVEL_10', 'GRAVEL_-1', 'LOG:3_1', 'LOG:3_-1', 'LOG:3_3', 'LOG:3_4', 'LOG_2_6', 'LOG_2_2', 'LOG_2_3', 'LOG_2_-1']

  getCollLevelssss(collUnlocked);
}

function getCollLevelssss(collUnlocked) {
  var collName;
  var collLevel;

  collUnlocked.forEach(x => {
    collName = x.match('(.*\_)')[0];
    collName = collName.slice(0, -1);

    collLevel = x.match(''); //need regex right here

    console.log(x + ' | ' + collName + ' = ' + collLevel);
  });
}

Answer 1

在您的目标中，以下方法如何？此示例脚本使用 split().

示例脚本：

var collUnlocked = ['GRAVEL_1', 'GRAVEL_3', 'GRAVEL_2', 'GRAVEL_5', 'GRAVEL_4', 'GRAVEL_10', 'GRAVEL_-1', 'LOG:3_1', 'LOG:3_-1', 'LOG:3_3', 'LOG:3_4', 'LOG_2_6', 'LOG_2_2', 'LOG_2_3', 'LOG_2_-1']
var res = collUnlocked.map(e => {
  var temp = e.split("_");
  var last = temp.pop();
  return [temp.join("_"), last];
});
console.log(res)

当此脚本为运行时，得到如下结果

  [ 
    [ 'GRAVEL', '1' ],
    [ 'GRAVEL', '3' ],
    [ 'GRAVEL', '2' ],
    [ 'GRAVEL', '5' ],
    [ 'GRAVEL', '4' ],
    [ 'GRAVEL', '10' ],
    [ 'GRAVEL', '-1' ],
    [ 'LOG:3', '1' ],
    [ 'LOG:3', '-1' ],
    [ 'LOG:3', '3' ],
    [ 'LOG:3', '4' ],
    [ 'LOG_2', '6' ],
    [ 'LOG_2', '2' ],
    [ 'LOG_2', '3' ],
    [ 'LOG_2', '-1' ]
  ]

参考：

split()

Answer 2

分裂与重组

function myfunk() {
  let a = ['GRAVEL_1', 'GRAVEL_3', 'GRAVEL_2', 'GRAVEL_5', 'GRAVEL_4', 'GRAVEL_10', 'GRAVEL_-1', 'LOG:3_1', 'LOG:3_-1', 'LOG:3_3', 'LOG:3_4', 'LOG_2_6', 'LOG_2_2', 'LOG_2_3', 'LOG_2_-1'];
  let b = a.reduce((a,c) => {
    let t = c.split('_');
    if(t.length == 2) {
      a.arr.push(`${t[0]},${t[1]}`);
    } else if(t.length == 3) {
      a.arr.push(`${t[0]}_${t[1]},${t[2]}`);
    }
    return a;
  },{arr:[]}).arr;
  Logger.log(JSON.stringify(b));
}

Execution log
7:10:50 PM  Notice  Execution started
7:10:50 PM  Info    ["GRAVEL,1","GRAVEL,3","GRAVEL,2","GRAVEL,5","GRAVEL,4","GRAVEL,10","GRAVEL,-1","LOG:3,1","LOG:3,-1","LOG:3,3","LOG:3,4","LOG_2,6","LOG_2,2","LOG_2,3","LOG_2,-1"]
7:10:51 PM  Notice  Execution completed

Answer 3

你可以拆分 negative look ahead:

const collUnlocked = ['GRAVEL_1', 'GRAVEL_3', 'GRAVEL_2', 'GRAVEL_5', 'GRAVEL_4', 'GRAVEL_10', 'GRAVEL_-1', 'LOG:3_1', 'LOG:3_-1', 'LOG:3_3', 'LOG:3_4', 'LOG_2_6', 'LOG_2_2', 'LOG_2_3', 'LOG_2_-1']
const res = collUnlocked.map(e => e.split(/_(?!.*_.*)/));
console.log(res)

/_(?!.*_.*) - _ 后面没有任何 .*_.*（_ 周围的任何字符）

正则表达式匹配第二个下划线后的所有内容

regex to match everything after second underscore

regex

google-apps-script

示例脚本：

参考：

分裂与重组