Question

我试图找到一种方法从数组中删除与给定模式匹配的一堆元素。

我的数组是：

var columns_array = [
   "table_a.id", 
   "table_a.name", 
   "table_a.surname", 
   "table_b.user_id", 
   "table_b.dob", 
   "table_b.start_date"
]

现在我希望从该数组中删除与给定模式匹配的所有内容：

var rm_matching = "table_b"

以便生成的数组看起来像这样：

var columns_array = [
   "table_a.id", 
   "table_a.name", 
   "table_a.surname"
]

我厌倦了拼接，地图和谷歌搜索提供的其他解决方案的少数变体（太多不能放在这里）但从未得到预期的结果。

我该如何处理？

Answer 1

我会选择filter功能：

var rm_matching = "table_b";
columns_array = columns_array.filter(
    function(elem) {
        return (!elem.match(rm_matching + "(.*)"));
    });

Answer 2

使用var columns_array = [ "table_a.id", "table_a.name", "table_a.surname", "table_b.user_id", "table_b.dob", "table_b.start_date" ] var column_a = columns_array.filter(a => /table_a/.test(a)); var list = document.createElement("ul"); document.body.appendChild(list); for (var i = 0; i < column_a.length; i++) { var listItem = document.createElement("li"); listItem.innerHTML = column_a[i]; document.body.appendChild(listItem); }

＆＃13;

myorg_customer

＆＃13;

Answer 3

试试这个

    var columns_array = [
        "table_a.id",
        "table_a.name",
        "table_b.start_date",
        "table_a.surname",
        "table_b.start_date",
        "table_b.user_id",
        "table_a.surname",
        "table_b.dob",
        "table_b.start_date"
    ];
    var str = "table_b"; // enter string you want to match
    var newArr = Array();
    var i = 0;
    columns_array.forEach(function(item, index)
    {
        if (item.indexOf(str) == -1)
        {
            newArr [i] = item;
            ++i;
        }
    });
    columns_array = newArr;
    delete newArr;

从数组中删除与模式匹配的项目

3 个答案: