首页 \ 问答 \ console.log(array)返回空,但console.log(array.length)不返回0(console.log(array) returns empty, but console.log(array.length) doesn't return 0)

console.log(array)返回空,但console.log(array.length)不返回0(console.log(array) returns empty, but console.log(array.length) doesn't return 0)

我有以下javascript

var filtered = [];

$('#footballCheck').on('change', function() {
  if ($('#footballCheck').is(':checked')) {
    for (var i = 0; i < fa_names.length; i++) {
      if (fa_names[i]["facility_activity"].toLowerCase().indexOf("football") >= 0) {
        filtered.push(fa_names[i]);
      }
    }

    filtered.sort(SortByName);
    $('#mCSB_1_container').empty(".facility-name");
    for (var k = 0; k < filtered.length; k++) {
      $('#mCSB_1_container').append('<div class="row facility-name">\
                                             <button class="btn btn-default btn-fill btn-menu" date-name="' + filtered[k]["facility_name"] + '">' + filtered[k]["facility_name"] + '</button></div>');
      $('#facilities-body').mCustomScrollbar("update");
    }
  } else {
    for (var j = 0; j < filtered.length; j++) {
      if (filtered[j]["facility_activity"].toLowerCase().indexOf("football") >= 0) {
        delete filtered[j];
      }
    }
    console.log(filtered);
    console.log(filtered.length);
    if (filtered.length > 0) {
      filtered.sort(SortByName);
      $('#mCSB_1_container').empty(".facility-name");
      for (var k = 0; k < filtered.length; k++) {
        $('#mCSB_1_container').append('<div class="row facility-name">\
                                                 <button class="btn btn-default btn-fill btn-menu" date-name="' + filtered[k]["facility_name"] + '">' + filtered[k]["facility_name"] + '</button></div>');
        $('#facilities-body').mCustomScrollbar("update");
      }
    } else {
      $('#mCSB_1_container').empty(".facility-name");
      for (k = 0; k < fa_names.length; k++) {
        $('#mCSB_1_container').append('<div class="row facility-name">\
                                                 <button class="btn btn-default btn-fill btn-menu" date-name="' + fa_names[k]["facility_name"] + '">' + fa_names[k]["facility_name"] + '</button></div>');
        $('#facilities-body').mCustomScrollbar("update");
      }
    }
  }
});

应该做以下事情:

  • 如果选中该复选框,则浏览另一个数组并将键值“facility_activity”包含“football”的所有对象推送到已filtered数组
  • 如果未选中,则从filtered数组中删除其键值“facility_activity”包含“football”的所有对象
  • 如果未选中并且其长度> 0(意味着仍有过滤的对象)则打印它们
  • 如果取消选中并且filtered数组为空,则追加父数组中的所有项(未过滤的项)

这是小提琴

背景:复选框充当搜索的filtered器, filtered数组存储那些过滤后的值(除了那个小提琴之外的其他地方, filtered数组还没有更改)。

我的问题是console.log(filtered)返回[]console.log(filtered.length)不返回0 ,它应该。 为什么会发生这种情况,我该如何解决?


I have the following javascript

var filtered = [];

$('#footballCheck').on('change', function() {
  if ($('#footballCheck').is(':checked')) {
    for (var i = 0; i < fa_names.length; i++) {
      if (fa_names[i]["facility_activity"].toLowerCase().indexOf("football") >= 0) {
        filtered.push(fa_names[i]);
      }
    }

    filtered.sort(SortByName);
    $('#mCSB_1_container').empty(".facility-name");
    for (var k = 0; k < filtered.length; k++) {
      $('#mCSB_1_container').append('<div class="row facility-name">\
                                             <button class="btn btn-default btn-fill btn-menu" date-name="' + filtered[k]["facility_name"] + '">' + filtered[k]["facility_name"] + '</button></div>');
      $('#facilities-body').mCustomScrollbar("update");
    }
  } else {
    for (var j = 0; j < filtered.length; j++) {
      if (filtered[j]["facility_activity"].toLowerCase().indexOf("football") >= 0) {
        delete filtered[j];
      }
    }
    console.log(filtered);
    console.log(filtered.length);
    if (filtered.length > 0) {
      filtered.sort(SortByName);
      $('#mCSB_1_container').empty(".facility-name");
      for (var k = 0; k < filtered.length; k++) {
        $('#mCSB_1_container').append('<div class="row facility-name">\
                                                 <button class="btn btn-default btn-fill btn-menu" date-name="' + filtered[k]["facility_name"] + '">' + filtered[k]["facility_name"] + '</button></div>');
        $('#facilities-body').mCustomScrollbar("update");
      }
    } else {
      $('#mCSB_1_container').empty(".facility-name");
      for (k = 0; k < fa_names.length; k++) {
        $('#mCSB_1_container').append('<div class="row facility-name">\
                                                 <button class="btn btn-default btn-fill btn-menu" date-name="' + fa_names[k]["facility_name"] + '">' + fa_names[k]["facility_name"] + '</button></div>');
        $('#facilities-body').mCustomScrollbar("update");
      }
    }
  }
});

that should do the following:

  • If the checkbox is checked, go through another array and push all objects whose key value "facility_activity" contains "football" to the filtered array
  • If it is unchecked, then remove all objects from filtered array whose key value "facility_activity" contains "football"
  • If it is unchecked and its length is > 0 (meaning that there are still filtered objects) print those
  • If it is unchecked and the filtered array is empty, then append all items from the parent array (the one that is not filtered)

Here is the fiddle.

Background: The checkboxes act as filters for a search and the filtered array stores those filtered values (in no other place other than what is in that fiddle the filtered array is changed yet).

My problem is that console.log(filtered) returns [] and console.log(filtered.length) does NOT return 0 and it should. Why does this happen and how can I fix it?


原文:https://stackoverflow.com/questions/36825511
更新时间:2023-02-10 08:02

最满意答案

您可以通过将一系列序列转换线程化来完成此操作。

(->> data
     (group-by #(->> % key (take 2)))
     vals 
     (map (comp first first (partial sort-by (comp - val))))
     (map (juxt #(subvec % 0 2) #(% 2)))
     (into {}))

;{[0 1] "a", [1 1] "a"}

......在哪里

(def data {[0 1 "a"] 2, [0 1 "b"] 1, [1 1 "a"] 1})

您逐行构建解决方案。 我建议你跟随建筑的脚步,从...开始

(->> data
     (group-by #(->> % key (take 2)))

;{(0 1) [[[0 1 "a"] 2] [[0 1 "b"] 1]], (1 1) [[[1 1 "a"] 1]]}

堆叠(懒惰)序列的层可以运行得相当慢,但是Clojure 1.7中提供的换能器将允许您用这个成语编写更快的代码,如这个优秀的答案所示


You can do this by threading together a series of sequence transformations.

(->> data
     (group-by #(->> % key (take 2)))
     vals 
     (map (comp first first (partial sort-by (comp - val))))
     (map (juxt #(subvec % 0 2) #(% 2)))
     (into {}))

;{[0 1] "a", [1 1] "a"}

... where

(def data {[0 1 "a"] 2, [0 1 "b"] 1, [1 1 "a"] 1})

You build up the solution line by line. I recommend you follow in the footsteps of the construction, starting with ...

(->> data
     (group-by #(->> % key (take 2)))

;{(0 1) [[[0 1 "a"] 2] [[0 1 "b"] 1]], (1 1) [[[1 1 "a"] 1]]}

Stacking up layers of (lazy) sequences can run fairly slowly, but the transducers available in Clojure 1.7 will allow you to write faster code in this idiom, as seen in this excellent answer.

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