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      1. 根据单独元素中的光标位置旋转元素

        时间:2023-10-13

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                  本文介绍了根据单独元素中的光标位置旋转元素的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

                  问题描述

                  I've been Working on a breadcrumbs directory feature recently that requires an element to rotate based on the cursor x position within the breadcrumbs container element. Long story short, I need the arrow in the lower '#pointer-box' to always point at the cursor when it's within the '#target-box'.

                  I'm looking for two separate formulas that will a.) set the initial left-most position of the arrow when the '#target-box' cursor x position is at 0, and b.) keep the arrow's left-most and right-most rotation properties proportional at any browser width or on window resize.

                  Any help is greatly appreciated.

                  Here's the live demo. http://jsfiddle.net/HeFqh/

                  Thank you

                  Update

                  With help from Tats_innit I was able to get the arrow pointing at the cursor when it's inside the '#target-box'. Now I have two specific issues to solve.

                  a.) When the window is resized the arrow and cursor are no longer aligned.

                  b.) The 'var y' on 'mousemove' is not deducting the top offset

                  var y = e.pageY - this.offsetTop
                  

                  The updated live demo. http://jsfiddle.net/HeFqh/11/

                  Thank you

                  解决方案

                  Hiya from @brenjt's :) response above pasting this as answer post & here is a sample demo http://jsfiddle.net/JqBZb/

                  Thanks and further helpful link here: jQuery resize to aspect ratio & here How to resize images proportionally / keeping the aspect ratio?

                  Please let me know if I missed anything! Hope this helps! have a nice one, Cheers!

                  jquery code

                  var img = $('.image');
                  if(img.length > 0){
                      var offset = img.offset();
                      function mouse(evt){
                          var center_x = (offset.left) + (img.width()/2);
                          var center_y = (offset.top) + (img.height()/2);
                          var mouse_x = evt.pageX; var mouse_y = evt.pageY;
                          var radians = Math.atan2(mouse_x - center_x, mouse_y - center_y);
                          var degree = (radians * (180 / Math.PI) * -1) + 90; 
                          img.css('-moz-transform', 'rotate('+degree+'deg)');
                          img.css('-webkit-transform', 'rotate('+degree+'deg)');
                          img.css('-o-transform', 'rotate('+degree+'deg)');
                          img.css('-ms-transform', 'rotate('+degree+'deg)');
                      }
                      $(document).mousemove(mouse);
                  }
                  
                  ​
                  

                  这篇关于根据单独元素中的光标位置旋转元素的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持html5模板网!

                  上一篇:使用 javascript 旋转 div 下一篇:如何获得用css旋转转换的元素的位置

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