字符串操作 JavaScript - 替换占位符

Question

我有一个很长的字符串，我必须以特定的方式对其进行操作。 该字符串可以包含导致我的代码出现问题的其他子字符串。 因此，在对字符串执行任何操作之前，我将所有子字符串（由"引入并以非转义"结尾的任何内容）替换为以下格式的占位符： $0 , $1 , $2 , ..., $n 。 我确信主字符串本身不包含字符$但子字符串之一（或更多）可能是例如"$0" 。

现在的问题是：在操作/格式化主字符串后，我需要再次用它们的实际值替换所有占位符。

方便地我将它们保存为这种格式：

// TypeScript
let substrings: { placeholderName: string; value: string }[];

但是做：

 // JavaScript let mainString1 = "main string $0 $1"; let mainString2 = "main string $0 $1"; let substrings = [ { placeholderName: "$0", value: "test1 $1" }, { placeholderName: "$1", value: "test2" } ]; for (const substr of substrings) { mainString1 = mainString1.replace(substr.placeholderName, substr.value); mainString2 = mainString2.replaceAll(substr.placeholderName, substr.value); } console.log(mainString1); // expected result: "main string test1 test2 $1" console.log(mainString2); // expected result: "main string test1 test2 test2" // wanted result: "main string test1 $1 test2"

不是一个选项，因为子字符串可能包含$x ，它会替换错误的东西（通过.replace()和通过.replaceAll() ）。

获取子字符串是用正则表达式存档的，也许正则表达式在这里也有帮助？ 虽然我无法控制子字符串中保存的内容......

Answer 1

如果您确定所有占位符都遵循$x格式，我会使用带有回调的.replace()方法：

const result = mainString1.replace(
  /\$\d+/g,
  placeholder => substrings.find(
    substring => substring.placeholderName === placeholder
  )?.value ?? placeholder
);

// result is "main string test1 $1 test2"

Answer 2

这可能不是最有效的代码。 但这是我用评论制作的功能。

注意：要小心，因为如果您将相同的占位符放入其内部，它将创建一个无限循环。 前任：

{ placeholderName: "$1", value: "test2 $1" }

 let mainString1 = "main string $0 $1"; let mainString2 = "main string $0 $1"; let substrings = [{ placeholderName: "$0", value: "test1 $1" }, { placeholderName: "$1", value: "test2" }, ]; function replacePlaceHolders(mainString, substrings) { let replacedString = mainString //We will find every placeHolder, the followin line wil return and array with all of them. Ex: ['$1', $n'] let placeholders = replacedString.match(/\\$[0-9]*/gm) //while there is some place holder to replace while (placeholders !== null && placeholders.length > 0) { //We will iterate for each placeholder placeholders.forEach(placeholder => { //extrac the value to replace let value = substrings.filter(x => x.placeholderName === placeholder)[0].value //replace it replacedString = replacedString.replace(placeholder, value) }) //and finally see if there is any new placeHolder inserted in the replace. If there is something the loop will start again. placeholders = replacedString.match(/\\$[0-9]*/gm) } return replacedString } console.log(replacePlaceHolders(mainString1, substrings)) console.log(replacePlaceHolders(mainString2, substrings))

编辑：

好的...我想我现在明白你的问题了...你不希望你的值中的 placeHoldersLike 字符串被替换。

这个版本的代码应该按预期工作，你不必担心这里的内嵌循环。 但是，请注意您的占位符，“$”是正则表达式中的保留字符，您应该避免使用它们。 我假设你所有的 placeHolders 都会像“$1”、“$2”等。如果不是，你应该编辑 regexPlaceholder 函数来包装和转义那个字符。

 let mainString1 = "main string $0 $1"; let mainString2 = "main string $0 $1 $2"; let substrings = [ { placeholderName: "$0", value: "$1 test1 $2 $1" }, { placeholderName: "$1", value: "test2 $2" }, { placeholderName: "$2", value: "test3" }, ]; function replacePlaceHolders(mainString, substrings) { //You will need to escape the $ characters or maybe even others depending of how you made your placeholders function regexPlaceholder(p) { return new RegExp('\\\\' + p, "gm") } let replacedString = mainString //We will find every placeHolder, the followin line wil return and array with all of them. Ex: ['$1', $n'] let placeholders = replacedString.match(/\\$[0-9]*/gm) //if there is any placeHolder to replace if (placeholders !== null && placeholders.length > 0) { //we will declare some variable to check if the values had something inside that can be //mistaken for a placeHolder. //We will store how many of them have we changed and replace them back at the end let replacedplaceholdersInValues = [] let indexofReplacedValue = 0 placeholders.forEach(placeholder => { //extrac the value to replace let value = substrings.filter(x => x.placeholderName === placeholder)[0].value //find if the value had a posible placeholder inside let placeholdersInValues = value.match(/\\$[0-9]*/gm) if (placeholdersInValues !== null && placeholdersInValues.length > 0) { placeholdersInValues.forEach(placeholdersInValue => { //if there are, we will replace them with another mark, so our primary function wont change them value = value.replace(regexPlaceholder(placeholdersInValue), "<markToReplace" + indexofReplacedValue + ">") //and store every change to make a rollback later replacedplaceholdersInValues.push({ placeholderName: placeholdersInValue, value: "<markToReplace" + indexofReplacedValue + ">" }) }) indexofReplacedValue++ } //replace the actual placeholders replacedString = replacedString.replace(regexPlaceholder(placeholder), value) }) //if there was some placeholderlike inside the values, we change them back to normal if (replacedplaceholdersInValues.length > 0) { replacedplaceholdersInValues.forEach(replaced => { replacedString = replacedString.replace(replaced.value, replaced.placeholderName) }) } } return replacedString } console.log(replacePlaceHolders(mainString1, substrings)) console.log(replacePlaceHolders(mainString2, substrings))

Answer 3

关键是选择一个在主字符串和子字符串中都不可能出现的占位符。 我的技巧是使用不可打印的字符作为占位符。 我最喜欢的是NUL字符（ 0x00 ），因为大多数其他人不会使用它，因为 C/C++ 认为它是字符串的结尾。 然而，Javascript 足够强大，可以处理包含 NUL 的字符串（编码为 un​​icode \\0000）：

let mainString1 = "main string \0-0 \0-1";
let mainString2 = "main string \0-0 \0-1";

let substrings = [
  { placeholderName: "\0-0", value: "test1 $1" },
  { placeholderName: "\0-1", value: "test2" }
];

其余代码不需要更改。

请注意，我使用-字符来防止 javascript 将您的数字0和1解释为八进制\\0 。

如果您像大多数程序员一样厌恶\\0那么您可以使用任何其他非打印字符，如\\1 （标题开头）、 007 （使您的终端发出铃声的字符 - 还有詹姆斯邦德）等。