How to select a range of elements in repeated pattern

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既然无缘
既然无缘 2020-11-22 08:16

Let\'s say there are 12 items in 4 rows.

 |  1   ||  2   ||  3   |
 |  4   ||  5   ||  6   |
 |  7   ||  8   ||  9   |
 |  10  ||  11  ||  12  |
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  • 2020-11-22 08:44

    This is a commonly-asked question, but I wanted to point out that the reason :nth-child(n+4):nth-child(-n+6) only matches one specific range of elements is that it only provides a single start point (n+4) and a single end point (-n+6). The only elements that can be greater than or equal to 4 and less than or equal to 6 are 4, 5 and 6, so it's impossible to match elements outside of this range using the same selector. Adding more :nth-child() pseudos will only narrow down the matches.

    The solution is to think of this in terms of columns, assuming there will always be exactly 3 columns (elements) per row. You have three columns, so you will need three separate :nth-child() pseudos. Elements 4 and 10 from the first column are 6 elements apart, so the argument to all of the :nth-child() pseudos needs to start with 6n.

    The +b portion in the An+B expression can either be +4, +5 and +6, or 0, -1 and -2 — they will both match the same set of elements:

    • li:nth-child(6n+4), li:nth-child(6n+5), li:nth-child(6n+6)
    • li:nth-child(6n), li:nth-child(6n-1), li:nth-child(6n-2)

    You cannot do this with a single :nth-child() pseudo-class, or a single compound selector consisting of any combination of :nth-child() pseudos, because the An+B notation simply doesn't allow such expressions to be constructed that match elements in ranges as described.

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  • 2020-11-22 09:01

    You can't do this with a single selector and need to use a grouping of nth-child selectors like in the below snippet:

    li:nth-child(6n), li:nth-child(6n-1), li:nth-child(6n-2) {
      color: red;
    }
    <ul>
      <li>1</li>
      <li>2</li>
      <li>3</li>
      <li>4</li>
      <li>5</li>
      <li>6</li>
      <li>7</li>
      <li>8</li>
      <li>9</li>
      <li>10</li>
      <li>11</li>
      <li>12</li>
    </ul>

    Selector Explanation:

    • li:nth-child(6n) - Selects the 6th (=6*1), 12th (=6*2), 18th (=6*3) elements and so on.
    • li:nth-child(6n-1) - Selects the 5th (=(6*1)-1), 11th (=(6*2)-1), 17th (=(6*3)-1) elements and so on.
    • li:nth-child(6n-2) - Selects the 4th (=(6*1)-2), 10th (=(6*2)-2), 16th (=(6*3)-2) elements and so on.

    Note: Generally n starts from 0 but I have excluded it in the explanation because we don't have any 0th, -1th or -2nd element.

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