如果之前没有另一个字符，如何查找字符?

I've got a string

我有一个字符串

{'lalala'} text before \{'lalala'\} {'lalala'} text after

I want to get open bracket { but only if there is no escape char \ before.

我想要获取左括号{但前提是之前没有转义字符。

Kind of /(?:[^\\])\{/ but it doesn't work at first statement.

/(?:[^ \ \])\ { /但它不工作首先声明。

2 个解决方案

#1

The typical approach is to match the non-\ preceding character (or beginning of string), and then put it back in your replacement logic.

典型的方法是匹配非\前的字符(或字符串的开头)，然后将其放回替换逻辑中。

const input = String.raw`{'lalala'} text before \{'lalala'\} {'lalala'} text after`;

function replace(str) {
  return input.replace(/(^|[^\\])\{'(\w+)'\}/g, 
    (_, chr, word) => chr + word.toUpperCase());
}

console.log(replace(input));

#2

That's where ^ comes in: it anchors a piece of regex to the start of the string (or the line, in m multiline mode). Because a 'valid' opening bracket is either at the start of a string or after a non-\ character, we can use the following regex:

^进来的地方:它锚的正则表达式的字符串(或线,米多行模式)。因为“有效”左括号要么在字符串的开头，要么在非\字符之后，我们可以使用以下regex:

/(?:^|[^\\])\{/g

I added the g global flag because we want to match all 'valid' opening brackets. Example use:

我添加了g全局标记，因为我们希望匹配所有“有效”的开始括号。使用示例:

console.log("{'lalala'} text before \\{'lalala'\\} {'lalala'} text after".match(/(?:^|[^\\])\{/g))

If you want to use the regex in a replace, you might want to capture the character before the bracket, as that gets replaced as well.

如果您希望在替换中使用regex，您可能希望在括号之前捕获字符，因为它也会被替换。

#1