如何使用Javascript將一個數組的字符串拆分成兩個組合

Question

我給出了一個字符串數組。它是一個帶括號的動作的文本。防爆。

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

最終的結果，我想是

actionArray = "           [waves] 

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

的actionArray將被隔開用於每個字符所以動作標籤將是對動作期間speechArray的頂部。我已經用下面的方法解決了這個問題。

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，爲了提取與動作支架字符串的一部分，我有一個，如果方法找到「[」如果確實如此，那將它添加到actionArray，與IF和增加保持actionArray直到找到結束標記「]」

它的工作原理，但計數器letter不會繼續在那裏我找到關閉標籤後離開，所以整個字符串最終還是被複制到speechArray。然後我最終得到像這樣的東西

actionArray = "           [waves] 

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

我是新來的Javascript，沒有使用指針我不知道如何處理這個。有人能指出我正確的方向嗎？謝謝！

Answer 1

您可以使用String.prototype.match()，並且String.prototype.replace()與RegExp/\[.+\]/匹配字符開頭內"["後跟一個或多個字符，然後和包括"]"，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

您可以檢查此簡單的代碼片段：

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) 

輸出應該是：

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

Answer 3

我有下面的實現，也許它可能是你合適

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

使用正則表達式的簡單解決方案。

首先最容易的部分（speechString）：

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

它只是空格數替換[...]等於比賽的長度。

現在較硬部分（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); 

它定義了相同的正則表達式找到[...]，但使用的索引（匹配的現在的位置）和匹配長度來構建的字符串等預期。

Here是一個工作小提琴。

---

EDIT actionString也可以由一個單一語句檢索：

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

匹配的所有字符和[...]表達並且在第一組的長度與P2所附的長度用空格替換它比賽。