javascript - 返回内容部分与模式匹配的第一个项目的 url

Question

如何在内容部分与pattern匹配的页面中返回第一页的url部分(函数1的风格，不区分大小写)，但如果没有找到页面则返回空字符串？

例如

url1(pages,"GREAT") returns "www.xyz.ac.uk"

url1(pages,"xyz") returns ""

这是迄今为止我的代码:

var pg = [ "|www.cam.ac.uk|Cambridge University offers degree programmes and world class research." , "!www.xyz.ac.uk!An great University" , "%www%Yet another University" ];
var pt = "great";

function url1(pages, pattern) {
    var result = "";
    for (x in pages) {
        current = pages[x].split(pattern);
        result = current[1];
    }
    return result;
}

alert(url1(pg, pt));

Answer 1

试试这个:

var pg = [ "|www.cam.ac.uk|Cambridge University offers degree programmes and world class research." , "!www.xyz.ac.uk!An great University" , "%www%Yet another University" ];

function find(pages, pattern) {
  var i, l, page, arr;
  pattern = pattern.toLowerCase();
  for(i=0, l=pages.length; i<l; i++) {
    page = pages[i];
    arr = page.split(page[0]);
    if(arr.slice(2).join(page[0]).toLowerCase().indexOf(pattern) >=0) {
      return arr[1];
    }
  }
  return '';
}

console.log(find(pg,'great')); // 'www.xyz.ac.uk'
console.log(find(pg,'xyz')); // ''

在这里摆弄:http://jsbin.com/uGebeQi/2/edit