按域和目录的Javascript组URL

时间：2017-02-19 14:08:16

标签： javascript arrays

如何按域和目录对排序列表中的URL进行分组？

  • 如果两个URL具有相同的目录（只是域后的第一个），则应将它们分组到一个数组中;

  • 第一个目录不同但域名相同的网址应按数组分组;

例如，此列表中的网址：

var url_list = ["https://www.facebook.com/impression.php/f2e61d9df/?lid=115",
"https://www.facebook.com/plugins/like.php?app_id=5",
"https://www.facebook.com/tr/a/?id=228037074239568",
"https://www.facebook.com/tr/b/?ev=ViewContent",
"http://www.marvel.com/abc?f=33",
"http://www.marvel.com/games?a=11",
"http://www.marvel.com/games?z=22",
"http://www.marvel.com/videos"]

应分组如下：

var group_url = [
    ["https://www.facebook.com/impression.php/f2e61d9df/?lid=115","https://www.facebook.com/plugins/like.php?app_id=5",],
    ["https://www.facebook.com/tr/a/?id=228037074239568","https://www.facebook.com/tr/b/?ev=ViewContent"],
    ["http://www.marvel.com/abc?f=33","http://www.marvel.com/videos"],
    ["http://www.marvel.com/games?a=11","http://www.marvel.com/games?z=22"]
]

我写了一些代码但只设法按域分组网址：

var group_url = [];
var count = 0;
var url_list = ["https://www.facebook.com/impression.php/f2e61d9df/?lid=115",
  "https://www.facebook.com/plugins/like.php?app_id=5",
  "https://www.facebook.com/tr/?id=228037074239568",
  "https://www.facebook.com/tr/?ev=ViewContent",
  "http://www.marvel.com/abc?f=33",
  "http://www.marvel.com/games?a=11",
  "http://www.marvel.com/games?z=22",
  "http://www.marvel.com/videos"]
      
for(i = 0; i < url_list.length; i++) {
  if(url_list[i] != "") {
    var current = url_list[i].replace(/.*?:\/\//g, "");
    var check = current.substr(0, current.indexOf('/'));
    group_url.push([])
    for(var j = i; j < url_list.length; j++) {
      var add_url = url_list[j];
      if(add_url.indexOf(check) != -1) {
        group_url[count].push(add_url);
        url_list[j] = "";
      }
      else {
        break;
      }
    }
    count += 1;
  }
}
    	
console.log(JSON.stringify(group_url));

5 个答案:

答案 0 :(得分：2)

您似乎希望按域+目录对URL进行分组，但如果它们最终在其组中单独存在，则只能按域重新组合这些URL。

为此您可以使用此脚本（ES5）：

var url_list = ["https://www.facebook.com/impression.php/f2e61d9df/?lid=115",
"https://www.facebook.com/plugins/like.php?app_id=5",
"https://www.facebook.com/tr/a/?id=228037074239568",
"https://www.facebook.com/tr/b/?ev=ViewContent",
"http://www.marvel.com/abc?f=33",
"http://www.marvel.com/games?a=11",
"http://www.marvel.com/games?z=22",
"http://www.marvel.com/videos"];

// Group the URLs, keyed by domain+dir
var hash = url_list.reduce(function (hash, url) {
    // ignore protocol, and extract domain and first dir:
    var domAndDir = url.replace(/^.*?:\/\//, '').match(/^.*?\..*?\/[^\/?#]*/)[0];
    hash[domAndDir] = (hash[domAndDir] || []).concat(url);
    return hash;
}, {});

// Regroup URLs by domain only, when they are alone for their domain+dir
Object.keys(hash).forEach(function (domAndDir) {
    if (hash[domAndDir].length == 1) {
        var domain = domAndDir.match(/.*\//)[0];
        hash[domain] = (hash[domain] || []).concat(hash[domAndDir]);
        delete hash[domAndDir];
    }
});
// Convert hash to array
var result = Object.keys(hash).map(function(key) {
    return hash[key];
});

// Output result
console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }

注意：我没有像你在评论中提到ES5那样使用ES6，但考虑使用ES6 Map来获得这样的哈希，这更适合这项工作。

答案 1 :(得分：0)

_renderItem: function (ul, item) {
        var t = '',
            result = '';
            searchedText = this.term;
        $.each(this.options.columns, function (index, column) {
        console.log(item);
            var reg = new RegExp('(' +searchedText+ ')', 'gi');
            t += '<span style="padding:0 4px;float:left;width:' + column.width + ';">' + item[column.valueField ? column.valueField : index].replace(reg, "<span style='color: red'>$1</span>") + '</span>'
        });
        result = $('<li></li>')
            .data('ui-autocomplete-item', item)
            .append('<a class="mcacAnchor">' + t + '<div style="clear: both;"></div></a>')
            .appendTo(ul);
        return result;
    }

像这样使用：

urllistextended=url_list.map(function(el){return el.split("://")[1].split("/");});//remove protocol, split by /

var obj={};
for(var index in urllistextended){
var el=urllistextended[index];
    obj[el[0]]=obj[el[0]]||{};
    obj[el[0]][el[1]]=obj[el[0]][el[1]]||[];
    obj[el[0]][el[1]].push(url_list[index]);
 }

http://jsbin.com/qacasexowi/edit?console输入“结果”

答案 2 :(得分：0)

我建议使用优秀的URI.js库，它提供了解析，查询和操作网址的绝佳方法：http://medialize.github.io/URI.js/

例如，要使用路径（您所指的目录），您可以轻松地执行以下操作（直接从api文档中获取）：

var uri = new URI("http://example.org/foo/hello.html");
// get pathname
uri.pathname(); // returns string "/foo/hello.html"
// set pathname
uri.pathname("/foo/hello.html"); // returns the URI instance for chaining

// will encode for you
uri.pathname("/hello world/");
uri.pathname() === "/hello%20world/";
// will decode for you
uri.pathname(true) === "/hello world/";

// will return empty string for empty paths, but:
URI("").path() === "";
URI("/").path() === "/";
URI("http://example.org").path() === "/";

其余的应该很容易。

答案 3 :(得分：0)

我建议按域使用对象和分组，然后使用域后的第一个字符串。然后迭代树并将其减少到所需的结构。

此解决方案适用于未分类的数据。

＆＃13;
＆＃13; 
var url_list = ["https://www.facebook.com/impression.php/f2e61d9df/?lid=115", "https://www.facebook.com/plugins/like.php?app_id=5", "https://www.facebook.com/tr/a/?id=228037074239568", "https://www.facebook.com/tr/b/?ev=ViewContent", "http://www.marvel.com/abc?f=33", "http://www.marvel.com/games?a=11", "http://www.marvel.com/games?z=22", "http://www.marvel.com/videos"],
    temp = [],
    result;

url_list.forEach(function (a) {
    var m = a.match(/.*?:\/\/([^\/]+)\/?([^\/?]+)?/);
    m.shift();
    m.reduce(function (r, b) {
        if (!r[b]) {
            r[b] = { _: [] };
            r._.push({ name: b, children: r[b]._ });
        }
        return r[b];
    }, this)._.push(a);
}, { _: temp });

result = temp.reduce(function (r, a) {
    var top = [],
        parts = [];

    a.children.forEach(function (b) {
        if (b.children.length === 1) {
            top.push(b.children[0]);
        } else {
            parts.push(b.children);
        }
    });
    return top.length ? r.concat([top], parts) : r.concat(parts);
}, []);

console.log(result);
＆＃13; 
.as-console-wrapper { max-height: 100% !important; top: 0; }
＆＃13;
＆＃13;
＆＃13;

答案 4 :(得分：0)

这完全符合您的需要：

var url_list = ["https://www.facebook.com/impression.php/f2e61d9df/?lid=115",
"https://www.facebook.com/plugins/like.php?app_id=5",
"https://www.facebook.com/tr/a/?id=228037074239568",
"https://www.facebook.com/tr/b/?ev=ViewContent",
"http://www.marvel.com/abc?f=33",
"http://www.marvel.com/games?a=11",
"http://www.marvel.com/games?z=22",
"http://www.marvel.com/videos"];

var folderGroups = {};
for (var i = 0; i < url_list.length; i++) {

  var myRegexp = /.*\/\/[^\/]+\/[^\/\?]+/g;
  var match = myRegexp.exec(url_list[i]);
  var keyForUrl = match[0];
  if (folderGroups[keyForUrl] == null) {
    folderGroups[keyForUrl] = [];
  }
  folderGroups[keyForUrl].push(url_list[i]);
}

var toRemove = [];
Object.keys(folderGroups).forEach(function(key,index) {
    if (folderGroups[key].length == 1) {
      toRemove.push(key);
    }
});
for (var i = 0; i < toRemove.length; i++) {
  delete folderGroups[toRemove[i]];
}

//console.log(folderGroups);

var domainGroups = {};
for (var i = 0; i < url_list.length; i++) {
 //Check if collected previously
  var myRegexpPrev = /.*\/\/[^\/]+\/[^\/\?]+/g;
  var matchPrev = myRegexpPrev.exec(url_list[i]);
  var checkIfPrevSelected = matchPrev[0];
  debugger;
  if (folderGroups[checkIfPrevSelected] != null) {
    continue;
  }
  //Get for domain group
  var myRegexp = /.*\/\/[^\/]+/g;
  var match = myRegexp.exec(url_list[i]);
  var keyForUrl = match[0];
  if (domainGroups[keyForUrl] == null) {
    domainGroups[keyForUrl] = [];
  }
  domainGroups[keyForUrl].push(url_list[i]);
}

//console.log(domainGroups);

var finalResult = {};
$.extend(finalResult, folderGroups, domainGroups);
console.log(Object.values(finalResult));
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

相关问题
最新问题