如何將基於數組元素的字符串拆分成數組，保留javascript中的拆分字

Question

我有一個字符串

sReasons =  "O9C2700021Not eligible for SDWCEOC3900015Service upgradeHJC3900015Service upgradeJ8C5000016Delivery Attempt";

我需要根據分隔符數組拆分上面的字符串

const separator = ["O9", "EO", "HJ", "J8"];

其中前 2 個字符 (O9) 代表網絡代碼，接下來 4 個另一個代碼 (C270) & 下 4 個字符 (0021) 不符合 SDWC 條件的字符串長度

其中分隔碼是唯一的，有2個大寫字母，除inEligType textMessage中不會重復

我需要創建一個格式為 json

{
    {inEligType: "O9", msgCode: "C270", msgLen: "0021", textMsg: "Not eligible for SDWC"},
    {inEligType: "EO", msgCode: "C390", msgLen: "0015", textMsg: "Service upgrade"},
    {inEligType: "HJ", msgCode: "C390", msgLen: "0015", textMsg: "Service upgrade"},
    {inEligType: "J8", msgCode: "C500", msgLen: "0016", textMsg: "Delivery Attempt"}
}

我基本上沒有根據給定的數組拆分字符串本身，我嘗試了以下

sReasons =  "O9C2700021Not eligible for SDWCEOC0900015Service upgradeHJC3900015Service upgradeJ8C5HJ0016Delivery Attempt";    
const separator = ["O9", "EO", "HJ", "J8"];

function formatReasons(Reasons: string) {
var words: any[] = Reasons.split(this.spearator); 
for(let word in words)
    {
       console.log(word) ;
    }
}
var result = formatReasons(sHdnReasonsCreate);
console.log("Returned Result: "+result);

但它給了我結果

["O9C2700021Not eligible for SDWCEOC0900015Service upgradeHJC3900015Service upgradeJ8C5HJ0016Delivery Attempt"]length: 1__proto__: Array(0)

Returned Address is: undefined

Answer 1

我的基於正則表達式的方法：

sReasons =  "O9C2700021Not eligible for SDWCEOC0900015Service upgradeHJC3900015Service upgradeJ8C5HJ0016Delivery Attempt";    
const separator = ["O9", "EO", "HJ", "J8"];

// build the regex based on separators
let regexPattern = '^';
separator.forEach(text => {
    regexPattern += `${text}(.*)`;
});
regexPattern += '$';

// match the reasons
let r = new RegExp(regexPattern);
let matches = sReasons.match(r);

// prepare to match each message
let msgMatcher = new RegExp('^(?<msgCode>.{4})(?<msgLen>.{4})(?<textMsg>.*)$');
let output = [];

for (let i=1; i<matches.length; i++) {
    // match the message
    const msg = matches[i].match(msgMatcher);

    // store
    let item = msg.groups;
    item.inEligType = separator[i-1];
    output.push(item);
}

console.log(JSON.stringify(output, null, 2));

生產

[
  {
    "msgCode": "C270",
    "msgLen": "0021",
    "textMsg": "Not eligible for SDWC",
    "inEligType": "O9"
  },
  {
    "msgCode": "C090",
    "msgLen": "0015",
    "textMsg": "Service upgrade",
    "inEligType": "EO"
  },
  {
    "msgCode": "C390",
    "msgLen": "0015",
    "textMsg": "Service upgrade",
    "inEligType": "HJ"
  },
  {
    "msgCode": "C5HJ",
    "msgLen": "0016",
    "textMsg": "Delivery Attempt",
    "inEligType": "J8"
  }
]

Answer 2

很可能textMsg字段或任何其他字段永遠不會包含您用於inEligType字段的兩個字母字符串。 但你絕對確定嗎？ 在我看來，數據格式確實希望有人通過特定長度的子字符串來解析它； 如果您可以根據分隔符進行拆分，為什么還要有一個msgLen字段？ 如果將來inEligType代碼列表發生變化怎么辦？

由於這些原因，我強烈建議您通過 substring 長度而不是分隔符匹配來解析。 這是一種可能的方法：

function formatReasons(reasons: string) {
  const ret = []
  while (reasons) {
    const inEligType = reasons.substring(0, 2);
    reasons = reasons.substring(2);
    const msgCode = reasons.substring(0, 4);
    reasons = reasons.substring(4);
    const msgLen = reasons.substring(0, 4);
    reasons = reasons.substring(4);
    const textMsg = reasons.substring(0, +msgLen);
    reasons = reasons.substring(+msgLen);
    ret.push({ inEligType, msgCode, msgLen, textMsg });
  }
  return ret;
}

您可以驗證它是否為您的示例sReasons字符串生成了預期的 output：

const formattedReasons = formatReasons(sReasons);
console.log(JSON.stringify(formattedReasons, undefined, 2));
/* [
  {
    "inEligType": "O9",
    "msgCode": "C270",
    "msgLen": "0021",
    "textMsg": "Not eligible for SDWC"
  },
  {
    "inEligType": "EO",
    "msgCode": "C090",
    "msgLen": "0015",
    "textMsg": "Service upgrade"
  },
  {
    "inEligType": "HJ",
    "msgCode": "C390",
    "msgLen": "0015",
    "textMsg": "Service upgrade"
  },
  {
    "inEligType": "J8",
    "msgCode": "C5HJ",
    "msgLen": "0016",
    "textMsg": "Delivery Attempt"
  }
] */

請注意，上面的實現不檢查字符串的格式是否正確； 現在，如果你把垃圾送進去，你就會把垃圾拿出來。 如果您想要更多的安全性，您可以進行運行時檢查並拋出錯誤，例如，意外地運行reasons字符串的末尾，或者找到不代表數字的msgLen字段。 並且可以進行重構，這樣就不會重復像const s = reasons.substring(0, n); reasons = reasons.substring(n)這樣的代碼。 const s = reasons.substring(0, n); reasons = reasons.substring(n) 。 但是基本算法就在那里。

Playground 代碼鏈接

Answer 3

RegExp 的另一種選擇，代碼更少

 // Your data const data = "O9C2700021Not eligible for SDWCEOC3900015Service upgradeHJC3900015Service upgradeJ8C5000016Delivery Attempt"; // Set your data splitters from array const spl = ["O9", "EO", "HJ", "J8"].join('|'); // Use regexp to parse data const results = []; data.replace(new RegExp(`(${spl})(\\w{4})(\\w{4})(.*?)(?=${spl}|$)`, 'g'), (m,a,b,c,d) => { // Form objects and push to res results.push({ inEligType: a, msgCode: b, msgLen: c, textMsg: d }); }); // Result console.log(results);

Answer 4

第一種方法，基於捕獲split消耗的正則表達式的組，由助手 function 處理，最后reduce d 減少到預期結果......

 function chunkRight(arr, chunkLength) { const list = []; arr = [...arr]; while (arr.length >= chunkLength) { list.unshift( arr.splice(-chunkLength) ); } return list; } // see also... [https://regex101.com/r/tatBAB/1] // with eg // (?<inEligType>O9|EO|HJ|J8)(?<msgCode>\w{4})(?<msgLen>\d{4}) //... or... // (O9|EO|HJ|J8)(\w{4})(\d{4}) // function extractStatusItems(str, separators) { const regXSplit = RegExp(`(${ separators.join('|') })(\\w{4})(\\d{4})`); const statusValues = String(str).split(regXSplit).slice(1); const groupedValues = chunkRight(statusValues, 4); return groupedValues.reduce((list, [inEligType, msgCode, msgLen, textMsg]) => list.concat({ inEligType, msgCode, msgLen, textMsg }), [] ); } const statusCode = 'O9C2700021Not eligible for SDWCEOC3900015Service upgradeHJC3900015Service upgradeJ8C5000016Delivery Attempt'; console.log( `statusCode... ${ statusCode }...`, extractStatusItems(statusCode, ['O9', 'EO', 'HJ', 'J8']) );

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...其次是第二種方法，幾乎完全基於捕獲命名組的正則表達式，由matchAll消耗，最后map進入預期結果...

 // see also... [https://regex101.com/r/tatBAB/2] // with eg // (?<inEligType>O9|EO|HJ|J8)(?<msgCode>\w{4})(?<msgLen>\d{4})(.*?)(?<textMsg>.*?)(?=O9|EO|HJ|J8|$) // function extractStatusItems(str, separators) { separators = separators.join('|'); const regXCaptureValues = RegExp( `(?<inEligType>${ separators })(?<msgCode>\\w{4})(?<msgLen>\\d{4})(.*?)(?<textMsg>.*?)(?=${ separators }|$)`, 'g' ); return [...String(str).matchAll(regXCaptureValues) ].map( ({ groups }) => ({...groups }) ); } const statusCode = 'O9C2700021Not eligible for SDWCEOC3900015Service upgradeHJC3900015Service upgradeJ8C5000016Delivery Attempt'; console.log( `statusCode... ${ statusCode }...`, extractStatusItems(statusCode, ['O9', 'EO', 'HJ', 'J8']) );

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