Question

我想指定要保留的元素，但要保留其结构。示例：

<html>
    <head>
        <meta xx>
    </head>
    <body>
        <div class="some1">
            <div class="some1-1"></div>
            <div class="some1-2"></div>
            <div class="some1-3"></div>
            <div class="some1-4"></div>
            <div class="some1-5"></div>
        </div>
        <div class="some2">
            <div class="some2-2">
                <div class="some2-2-1"></div>
                <div class="some2-2-2"></div>
            </div>
            <div class="some2-3"></div>
            <div class="some2-4"></div>
            <div class="some2-5"></div>
        </div>
    </body>
</html>

例如，我只需要保持： .some2-2-1，.some-2-3，.some-1-3，删除所有其他元素，并保留其父元素，这样它将是：

<html>
    <head>
        <meta xx>
    </head>
    <body>
        <div class="some1">
            <div class="some1-3"></div>
        </div>
        <div class="some2">
            <div class="some2-2">
                <div class="some2-2-1"></div>
            </div>
            <div class="some2-3"></div>
        </div>
    </body>
</html>

我不能指定要删除的元素（它是动态的），而只能指定要保留的元素，如何实现？

$('*:not('.some2-2-1, .some-2-3, .some-1-3')').remove() 在那种情况下是行不通的；）

Answer 1

我敢肯定会有几种方法，但这是一种方法（请参阅评论）：

$(function() {
    var elementsToKeep = [
        ".some2-2-1",
        ".some2-3",
        ".some1-3",
    ];

    // parent elements that are needed will be appended here
    var needed = [];

    for (var i = 0; i < elementsToKeep.length; i++) {
        // get the hierarchy, e.g. ".some2-2-1" -> "2-2-1"
        var hierarchyPos = elementsToKeep[i].replace(".some", "");

        // if we're keeping 2-2-1, we also need 2-2, and 2
        //
        // recursively remove the last two characters to get
        // the parent level until there's nothing left
        while (hierarchyPos !== "") {
            if (!needed.includes(hierarchyPos)) {
                needed.push(hierarchyPos);
            }

            hierarchyPos = hierarchyPos.slice(0, -2);
        }
    }

    // "needed" will now contain 2-2-1, 2-2, 2, etc.
    console.log("all required elements:", needed.join(", "));

    // iterate through each of the original selectors
    $(elementsToKeep.join(",")).each(function() {

        // check each of this element's siblings and remove
        // any whose class isn't included in the "needed" array
        var siblings = $(this).siblings();

        siblings.each(function(i, el) {
            var pos = el.className.replace("some", "");

            // if "pos" is 2-2-2 (for example), it won't appear in "needed",
            // so it's safe to remove
            if (!needed.includes(pos)) {
                $(el).remove();
            }
        })
    })

    // finished
    console.log("final output:");
    console.log($("#test").html());
})

<div id="test">
    <div class="some1">
        <div class="some1-1"></div>
        <div class="some1-2"></div>
        <div class="some1-3"></div>
        <div class="some1-4"></div>
        <div class="some1-5"></div>
    </div>
    <div class="some2">
        <div class="some2-2">
            <div class="some2-2-1"></div>
            <div class="some2-2-2"></div>
        </div>
        <div class="some2-3"></div>
        <div class="some2-4"></div>
        <div class="some2-5"></div>
    </div>
</div>

<script src="https://code.jquery.com/jquery-3.4.1.min.js"></script>

删除所有元素，但使用Cheerio / jQuery选择结构

1 个答案: