如何使用Javascript将一个数组的字符串拆分成两个组合

Question

我给出了一个字符串数组。它是一个带括号的动作的文本。防爆。

"John sees his neighbor and greets 'Hello!' [waves]. But she didn't hear him." 

最终的结果，我想是

actionArray = "           [waves] 

speechArray = "John sees his neighbor and greets 'Hello!'.  But she didn't hear him." 

的actionArray将被隔开用于每个字符所以动作标签将是对动作期间speechArray的顶部。我已经用下面的方法解决了这个问题。

var array = data.toString().split("\n"); 
    for(i in array) { 
    var actionArray = ""; 
    var speechArray = ""; 
    for (var letter in array[i]){ 
     if (array[i][letter] == "["){ 
     actionArray += array[i][letter]; 
     letter++; 
     while(array[i][letter] !== "]"){ 
      actionArray += array[i][letter]; 
      letter++; 
     } 
     if (array[i][letter] == "]"){ 
      actionArray += array[i][letter]; 
     } 

     } 

     if (array[i][letter] !== "["){ 

     speechArray += array[i][letter]; 
     actionArray += " "; 
     } 

因此，为了提取与动作支架字符串的一部分，我有一个，如果方法找到“[”如果确实如此，那将它添加到actionArray，与IF和增加保持actionArray直到找到结束标记“]”

它的工作原理，但计数器letter不会继续在那里我找到关闭标签后离开，所以整个字符串最终还是被复制到speechArray。然后我最终得到像这样的东西

actionArray = "           [waves] 

speechArray = "John sees his neighbor and greets 'Hello!'.[waves] But she didn't hear him." 

我是新来的Javascript，没有使用指针我不知道如何处理这个。有人能指出我正确的方向吗？谢谢！

Answer 1

您可以使用String.prototype.match()，并且String.prototype.replace()与RegExp/\[.+\]/匹配字符开头内"["后跟一个或多个字符，然后和包括"]"，String.prototype.repeat()

let str = "John sees his neighbor and greets 'Hello!' [waves]. But she didn't hear him."; 
 

 
let re = /\[.+\]/; 
 

 
let match = str.match(re); 
 

 
let res = " ".repeat(match.index) + match[0] 
 
      + " ".repeat(str.length - (match.index + match[0].length)); 
 

 
str = str.replace(re, " ".repeat(match[0].length)); 
 

 
let pre = document.querySelector("pre"); 
 

 
pre.textContent = res + "\n" + str;

<pre></pre>

Answer 2

您可以检查此简单的代码片段：

var data = "John sees his neighbor and greets 'Hello!' [waves]. But she didn't hear him." 
var arr = data.split(""); 
var action = false; 
var actionstr="",speachstr=""; 
    for(var i=0;i<arr.length;i++){ 
     if(arr[i]=='['){ 
      action = true 
     }else if(arr[i] ==']'){ 

      actionstr = actionstr.concat(arr[i]) 
      speachstr = speachstr.concat(" ") 
      action = false 
      continue 
     } 
     if(action){ 
      actionstr = actionstr.concat(arr[i]) 
      speachstr = speachstr.concat(" ") 
     }else{ 
      actionstr = actionstr.concat(" ") 
      speachstr = speachstr.concat(arr[i]) 
     } 
    } 
    console.log(actionstr) 
    console.log(speachstr) 

输出应该是：

          [waves]       
John sees his neighbor and greets 'Hello!'  . But she didn't hear him. 

Answer 3

我有下面的实现，也许它可能是你合适

var arr = "John sees his neighbor and greets 'Hello!' [waves]. But she didn't hear him." 

var p = arr.split(/\s+/) 

var c = p.map(value => { 

    if (/^\[.*\]/.test(value)) 
     return value; 
    else 
     return " " 
}) 

var command = c.join(" "); 
var strings = arr.replace(/\[.*\]/, " ".repeat(command.match(/\w+/)[0].length)) 


console.log(command) 
console.log(strings) 

Answer 4

使用正则表达式的简单解决方案。

首先最容易的部分（speechString）：

var speechString = string.replace(/(\[[^\]]*\])/gi, (v)=>' '.repeat(v.length)); 

它只是空格数替换[...]等于比赛的长度。

现在较硬部分（actionString）：

var actionString = ((string)=>{ 
    var ret = ''; 
    var re = /(\[[^\]]*\])/gi; 
    var l = 0; 
    while((m = re.exec(string)) != null) { 
     ret += ' '.repeat(m.index-l) + string.substring(m.index, m.index+m[1].length); 
     l = m.index+m[1].length; 
    } 
    return ret; 
})(string); 

它定义了相同的正则表达式找到[...]，但使用的索引（匹配的现在的位置）和匹配长度来构建的字符串等预期。

Here是一个工作小提琴。

---

EDIT actionString也可以由一个单一语句检索：

var actionString = string.replace(/([^\[]*)(\[[^\]]+?\])?/g, (v,p1,p2)=>{ 
    return (p1?' '.repeat(p1.length):'')+(p2||''); 
}); 

匹配的所有字符和[...]表达并且在第一组的长度与P2所附的长度用空格替换它比赛。