Question

假设我有一个包含两行的文件，我在文件中得到indexOf个子字符串。它返回18找到子字符串的字符。如何找到包含此信息的行？

var file = [
  'var foo = "hello"',
  'console.log(foo)',
].join('\n')

var char = file.indexOf('console') // => 18
var line = lineOfChar(file, char) // => 2

Answer 1

实现这一目标的一种可能方法是找到与您一样的字符串：

var index = file.indexOf('console'); // => 18

然后使用此索引创建一个包含该索引之前所有内容的子字符串：

var tempString = str.substring(0, index);

最后我们计算\n：

的出现次数

var lineNumber = tempString.split('\n').length;
// You should do - 1 if you want your 'first' line to be 0

Answer 2

function lineOf(text, substring){
  var line = 0, matchedChars = 0;

  for (var i = 0; i < text.length; i++) {
    text[i] === substring[matchedChars] ? matchedChars++ : matchedChars = 0;

    if (matchedChars === substring.length){
        return line;                  
    }
    if (text[i] === '\n'){
        line++;
    }
  }

  return  -1;
}

避免迭代字符串两次，一次找到子字符串并再次找到换行符。

Answer 3

var assert = require('assert')

var file = [
  'var alpha = "hello"',
  'var beta = "hello"',
  'var gamma = "hello"',
  'var delta = "hello"',
  'var episilon = "hello"'
].join('\n')

function getLine (body, charOrString) {
  if (!body) return false
  if (!charOrString) return false
  var char = (typeof charOrString === 'string') ? body.indexOf(charOrString) : charOrString
  var subBody = body.substring(0, char)
  if (subBody === '') return false
  var match = subBody.match(/\n/gi)
  if (match) return match.length + 1
  return 1
}

assert.equal(getLine(file, 'missing'), false)
assert.equal(getLine(file, 'alpha'), 1)
assert.equal(getLine(file, 'beta'), 2)
assert.equal(getLine(file, 'gamma'), 3)
assert.equal(getLine(file, 'delta'), 4)
assert.equal(getLine(file, 'episilon'), 5)

从文本正文中获取字符串的行号

3 个答案: