Question

需要定期表达。如果我想将“abcdf”改为“PPP”。

XabcdfX is xxxxxxx. <MabcdfM> has xxxxx. abcdf is xxxxxxx. <FabcdfF> zxabcdf abcdf.

然后是预期产出

XPPPX is xxxxxxx. <MabcdfM> has xxxxx. PPP is xxxxxxx. <FabcdfF> zxPPP PPP.

abcdf可以是整个扼杀物,甚至可以是一种扼杀。这只是字母数字。只有在不相隔的情况下才能取而代之。

我需要用javascript取而代之。请帮助我。

Answer 1

https://developer.mozilla.org/en/Java/Reference/Global_Objects/String/replace#Specification_a____as_a_paraile”rel=“nofollow” 查询replace:

s = s.replace(/abcdf|(<[^>]*>)/g, function(g0,g1){return g1 ? g1:  PPP ;});

Working Example:

显然,如果你把安格尔的括号nes为外壳,这将失败。

Answer 2

假设你的“主角”不能被推崇,你可以使用头指:

s.replace(/abcdf(?=[^<>]*(<|$))/g, "PPP")

Answer 3

如果你可以肯定,那就没有任何类似和大的;我会利用指数Of-功能,而不是使用一种管理:

index = string.indexOf("<", lastIndex);
index = string.indexOf(">", index);
index = string.indexOf("abcd", index);
string = string.substring(0, index)+"PPP"+string.substring(index+4);

Answer 4

var collect = function (s, re)
{
  var rv = [];
  for (var m; m = re.exec(s); rv.push(m));
  return rv;
}
var s = "XabcdfX is xxxxxxx. <MabcdfM> has xxxxx. abcdf is xxxxxxx. <FabcdfF> zxabcdf abcdf."
var tags = collect(s, /<[^>]*>/g);
var matches = collect(s, /abcdf/g);

console.log("INPUT:", s)

var r = s

MATCH: for (var i = 0, j = 0; i < matches.length; ++i) {
  var m = matches[i]
  var mbeg = m.index
  var mend = mbeg + m[0].length
  TAG: for (; j < tags.length; ++j) {
    var t = tags[j]
    var tbeg = t.index
    var tend = tbeg + t[0].length
    if (tend < mbeg)
      continue TAG
    if (tbeg < mbeg && tend > mend)
      continue MATCH
    if (tbeg > mbeg)
      break TAG
  }
  r = r.substr(0, mbeg) + m[0].toUpperCase() + r.substr(mbeg + m[0].length)
}
console.log("RESULT:", r)

友情链接