const charSequence = '=$A12+A$345+A6789';
const numberList = (charSequence
.split(/\$\d+/) // - split at "'$' followed by one or more numbers".
.join('') // - join array of split results into string again.
.match(/\d+/g) || []) // - match any number-sequence or fall back to empty array.
.map(str => +str); // - typecast string into number.
//.map(str => parseInt(str, 10)); // parse string into integer.
console.log('numberList : ', numberList);
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@ibraheem你能再次幫助我嗎？如果我想得到以下結果[“$A 13”，“A6790”]，如何增加ref輸出？- 揚·齊特尼亞克 23分鐘前
...該方法可以非?？焖俚氐?，因此它被證明是相當靈活的。split/join/match
const charSequence = '=$A13+A$345+A6790';
const numberList = (charSequence
.split(/\$\d+/) // - split at "'$' followed by one or more numbers".
.join('') // - join array of split results into string again.
.match(/\$*[A-Z]\d+/g) || []); // - match any sequence of an optional '$' followed
// by 1 basic latin uppercase character followed
// by one or more number character(s).
console.log('numberList : ', numberList);
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彼得感謝您對增量的快速響應，但在開始時，我有常量字符序列=“=$A 12 + A $ 345 + A6789”;并作為輸出，我需要[“$A 13”，“A6790”]。- 揚齊特尼亞克
...好吧，最后一個人將全面了解整個問題...這是（1）擺脫不必要的模式...（2）在特定模式中匹配數(shù)字，并以某種方式記住后者（3）增加這些數(shù)字，并以某種方式將它們重新設計成記憶/可回憶的模式。
const anchorSequence = '=$A12+A$345+A6789';
const listOfIncrementedAnchorCoordinates = [...(anchorSequence
// - split at "'$' followed by one or more numbers".
.split(/\$\d+/)
// - join array of split results into string again.
.join('')
// - match any sequence of an optional '$' followed by 1 basic latin
// uppercase character followed by one or more number character(s)
// and store each capture into a named group.
.matchAll(/(?<anchor>\$*[A-Z])(?<integer>\d+)/g) || [])
// map each regexp result from a list of RegExpStringIterator entries.
].map(({ groups }) => `${ groups.anchor }${ (+groups.integer + 1) }`);
console.log('listOfIncrementedAnchorCoordinates : ', listOfIncrementedAnchorCoordinates);
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彼得，如果你是感興趣的（...ed）在另一個問題中，我有一個。如何更改常量錨點序列 = '=$A 12+A$345+A6789';到以下輸出 [“B$345”，“B6789”]？我的意思是，如果字母不以$開頭，則按字母順序將字母更改為下一個字母（如果是A，則更改為B，如果是B，則更改為C，依此類推）。在我的例子中，它應該只改變A$345和A6789。- 揚齊特尼亞克
...通過一些思考努力，迭代/重構之前的版本到最后一個版本并不難......
const anchorSequence = '=$A12+A$345+A6789';
const listOfIncrementedColumns = [...(anchorSequence
// - split at "'$' followed by 1 basic latin uppercase character
// followed by one or more number character(s)".
.split(/\$[A-Z]\d+/)
// - join array of split results into string again.
.join('')
// - match any sequence of 1 basic latin uppercase character
// followed by an optional '$' followed by one or more number
// character(s) and store each capture into a named group.
.matchAll(/(?<column>[A-Z])(?<anchor>\$*)(?<row>\d+)/g) || [])
// map each regexp result from a list of RegExpStringIterator entries.
].map(({ groups }) => [
// - be aware that "Z" (charCode:90) will be followed by "[" (charCode:91)
// - thus, the handling of this edge case still needs to be implemented.
String.fromCharCode(groups.column.charCodeAt(0) + 1),
groups.anchor,
groups.row
].join(''));
console.log('listOfIncrementedColumns : ', listOfIncrementedColumns);
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