jQuery计算图像背景位置顶部位置？

Question

我将我的图像设置在CSS中心：

background-position: center center;

但是我想知道它在设置为中心后的顶级位置，我怎么能用jQuery或只是普通的javaScript来计算？

.parallax3 {
  position: relative;
  background-position: center center;
  background-repeat: no-repeat;
  background-size: cover;
  background-image: url('https://picsum.photos/g/200/600');
  height: 100%;
}

.tagline {
    font-size: 2.5vw;
    letter-spacing:0.05em;
    padding: 25% 25%;
}

.grid-x {
    border:1px solid red;
}

.cell {
    border: 1px solid blue;
}

<div class="row widescreen">

  <div class="grid-container full">
    <div class="grid-x grid-padding-x align-stretch small-padding-collapse">
      <div class="medium-6 cell">
        <div class="tagline text-center">
          <p>Lorem ipsum dolor sit amet</p>
        </div>
      </div>
      <div class="medium-6 cell parallax-viewport">
        <div class="parallax3">
        </div>
      </div>
    </div>
  </div>

</div>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.1/jquery.min.js"></script>
<link href="https://cdnjs.cloudflare.com/ajax/libs/foundicons/3.0.0/foundation-icons.css" rel="stylesheet">
    <link rel="stylesheet" href="https://cdnjs.cloudflare.com/ajax/libs/foundation/6.4.1/css/foundation.min.css">
    <script src="https://cdnjs.cloudflare.com/ajax/libs/foundation/6.4.1/js/foundation.min.js"></script>

有任何想法吗？

Answer 1

正如我在上面评论的那样，我们需要找到公式，因此我们的想法是依靠background-position:center和background-size:cover的逻辑。因此，由于图像将始终被拉伸，我们有两种可能性：

1. 图像的顶部将与div的顶部相同，因此top = 0
2. 图像的高度将比div长，顶部将被隐藏，因此顶部<0

你可能会注意到我认为相对于div顶部的顶部位置是我的起源

现在我们需要考虑图像如何覆盖div的所有不同情况，并根据每种情况计算图像的新高度。

这是一个代码示例：

//am using fixed values because they are known in this case, so this need to also be calculated
var him = 600;
var wim = 200;
var ratio = him / wim;
/*--*/

function get_top() {
  var h = $('.parallax3').height();
  var w = $('.parallax3').width();

  var diffh = h - him;
  var diffw = w - wim;
  if (diffw > 0 && diffh > 0) {
    //the image is smaller than the div so it should get bigger by at least the max difference
    if (diffh > diffw*ratio) {
      //both height will be equal here so top position = 0
      console.log(0);
    } else {
      //height of image will be bigger so top position < 0
      var newH = (wim + diffw) * ratio;
      console.log((h - newH) / 2)
    }
  } else if (diffw > 0) {
    //only the width if div is bigger so the image width will stretch and the height will be bigger and top < 0;
    var newH = (wim + diffw) * ratio;
    console.log((h - newH) / 2)
  } else if (diffh > 0) {
    //only the height is bigger so the image height will stretch and both heights will be equal to div and top = 0
    console.log(0);
  } else {
    // the image is bigger in this case so we do same logic as the start with abs value but we will reduce size so we consider the lowest value
    if (Math.abs(diffh) < Math.abs(diffw)*ratio) {
      //both height will be equal here so top position = 0
      console.log(0);
    } else {
      //height of image will remain bigger so top position < 0
      var newH = (wim + diffw) * ratio;
      console.log((h - newH) / 2)
    }

  }
}

get_top();

$(window).resize(function() {
  get_top()
})

body {
  height: 100vh;
}

.parallax3 {
  position: relative;
  background-position: center center;
  background-repeat: no-repeat;
  background-size: cover;
  background-image:url('https://picsum.photos/g/200/600');
  height: 100%;
}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="parallax3">
</div>

如果我们考虑在background-size中包含as值，则图像将始终包含在div中，因此两种可能性是：

1. 图像的顶部将与div的顶部相同，因此top = 0
2. 图像的高度会更小，图像的顶部将是可见的，因此顶部> 0

与上述相同，我们需要考虑不同的情况并计算图像的新高度。

这是一个代码示例：

//am using fixed values because they are known in this case, so this need to also be calculated
var him = 600;
var wim = 200;
var ratio = him / wim;
/*--*/

function get_top() {
  var h = $('.parallax3').height();
  var w = $('.parallax3').width();
  var diffh = h - him;
  var diffw = w - wim;
  if (diffw > 0 && diffh > 0) {
    //the image is smaller than the div so it should get bigger by at least the min difference
    if (diffh < diffw*ratio) {
      //both height will be equal here so top position = 0
      console.log(0);
    } else {
      //height of image will be bigger so top position < 0
      var newH = (wim + diffw) * ratio;
      console.log((h - newH) / 2)
    }
  } else if (diffw > 0) {
    //only the width if div is bigger so the image height need to be reduced and both height will be equal
    console.log(0);

  } else if (diffh > 0) {
    //only the height is bigger so the image width will be reduced 
    var newH = (wim + diffw) * ratio;
    console.log((h - newH) / 2);
  } else {
    // the image is bigger in this case so we do same logic as the start with abs value but we will reduce size so we consider the biggest value
    if (Math.abs(diffh) > Math.abs(diffw)*ratio) {
      //both height will be equal here so top position = 0
      console.log(0);
    } else {
      //height of image will be bigger so top position < 0
      var newH = (wim + diffw) * ratio;
      console.log((h - newH) / 2)
    }

  }
}

get_top();

$(window).resize(function() {
  get_top()
})

body {
  height: 100vh;
}

.parallax3 {
  position: relative;
  background-position: center center;
  background-repeat: no-repeat;
  background-size: contain;
  background-image: url('https://picsum.photos/g/200/600');
  height: 100%;
}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="parallax3">
</div>

两个代码段都可以进行优化以减少冗余，但我保留这些代码以更好地解释不同的情况。