How to find the highest z-index using jQuery

风流意气都作罢 提交于 2019-12-17 06:35:10

问题


I have a number of div elements with different z-index. And I want to find the highest z-index among these divs - how can I achieve it?

CSS:

#layer-1 { z-index: 1 }
#layer-2 { z-index: 2 }
#layer-3 { z-index: 3 }
#layer-4 { z-index: 4 }

HTML:

<div id="layer-1">layer-1</div>
<div id="layer-2">layer-2</div>
<div id="layer-3">layer-3</div>
<div id="layer-4">layer-4</div>

I don't think this line can find the highest z-index though.

var index_highest = parseInt($("div").css("zIndex"));
// returns 10000

回答1:


Note that z-index only affects positioned elements. Therefore, any element with position: static will not have a z-index, even if you assign it a value. This is especially true in browsers like Google Chrome.

var index_highest = 0;   
// more effective to have a class for the div you want to search and 
// pass that to your selector
$("#layer-1,#layer-2,#layer-3,#layer-4").each(function() {
    // always use a radix when using parseInt
    var index_current = parseInt($(this).css("zIndex"), 10);
    if(index_current > index_highest) {
        index_highest = index_current;
    }
});

JSFiddle demo

A general jQuery selector like that when used with an option that returns one value will merely return the first So your result is simply the z-index of the first div that jQuery grabs. To grab only the divs you want, use a class on them. If you want all divs, stick with div.




回答2:


Here is a very concise method:

var getMaxZ = function(selector){
    return Math.max.apply(null, $(selector).map(function(){
        var z;
        return isNaN(z = parseInt($(this).css("z-index"), 10)) ? 0 : z;
    }));
};

Usage:

getMaxZ($("#layer-1,#layer-2,#layer-3,#layer-4"));

Or, as a jQuery extension:

jQuery.fn.extend({
    getMaxZ : function(){
        return Math.max.apply(null, jQuery(this).map(function(){
            var z;
            return isNaN(z = parseInt(jQuery(this).css("z-index"), 10)) ? 0 : z;
        }));
    }
});

Usage:

$("#layer-1,#layer-2,#layer-3,#layer-4").getMaxZ();



回答3:


Besides @justkt's native solution above, there is a nice plugin to do what you want. Take a look at TopZIndex.

$.topZIndex("div");



回答4:


Try this :

var index_highest = 0;
$('div').each(function(){
    var index_current = parseInt($(this).css("z-index"), 10);
    if(index_current > index_highest) {
        index_highest = index_current;
    }
}); 



回答5:


This would do it:

$(document).ready(function() {
    var array = [];
    $("div").each(function() {
        array.push($(this).css("z-index"));
    });
    var index_highest = Math.max.apply(Math, array);
    alert(index_highest);
});

Try this




回答6:


I don't know how efficient this is, but you can use $.map to get all the z-indices:

var $divs = $('div'),
    mapper = function (elem) {
        return parseFloat($(elem).css('zIndex'));
    },
    indices = $.map($divs, mapper);

The indices variable is now an array of all the z-indices for all the divs. All you'd have to do now is apply them to Math.max:

var highest = Math.max.apply(whatevs, indices);



回答7:


This is taken directly from jquery-ui, it works really well:

(function ($) {
  $.fn.zIndex = function (zIndex) {
      if (zIndex !== undefined) {
        return this.css("zIndex", zIndex);
      }

      if (this.length) {
        var elem = $(this[ 0 ]), position, value;
        while (elem.length && elem[ 0 ] !== document) {
          // Ignore z-index if position is set to a value where z-index is ignored by the browser
          // This makes behavior of this function consistent across browsers
          // WebKit always returns auto if the element is positioned
          position = elem.css("position");
          if (position === "absolute" || position === "relative" || position === "fixed") {
            // IE returns 0 when zIndex is not specified
            // other browsers return a string
            // we ignore the case of nested elements with an explicit value of 0
            // <div style="z-index: -10;"><div style="z-index: 0;"></div></div>
            value = parseInt(elem.css("zIndex"), 10);
            if (!isNaN(value) && value !== 0) {
              return value;
            }
          }
          elem = elem.parent();
        }
      }

      return 0;
    }
})(jQuery);



回答8:


Here how I got both lowest/highest z-indexes. If you only want to get the highest z-index and nothing more, then this function may not efficient, but if you want to get all z-indexes and the ids associated with it (i.e. for use with bring 'layer' to front/send to back, bring forward, send backward, etc), this is one way to do it. The function returns an array of objects containing ids and their z-indexes.

function getZindex (id) {

     var _l = [];
     $(id).each(function (e) {
         // skip if z-index isn't set 
         if ( $(this).css('z-index') == 'auto' ) {
              return true
         }
         _l.push({ id: $(this), zindex: $(this).css('z-index') });
     });
     _l.sort(function(a, b) { return a.zindex - b.zindex });
     return _l;
}

// You'll need to add a class 'layer' to each of your layer
var _zindexes = getZindex('.layer');
var _length = _zindexes.length;

// Highest z-index is simply the last element in the array
var _highest = _zindexes[_length - 1].zindex

// Lowest z-index is simply the first element in the array
var _lowest = _zindex[0].zindex;

alert(_highest);
alert(_lowest);



回答9:


Vanilla JS, not 100% cross-browser. Including as reference for future readers/alternative method.

function getHighIndex (selector) {
    // No granularity by default; look at everything
    if (!selector) { selector = '*' };

    var elements = document.querySelectorAll(selector) ||
                   oXmlDom.documentElement.selectNodes(selector),
        i = 0,
        e, s,
        max = elements.length,
        found = [];

    for (; i < max; i += 1) {
        e = window.getComputedStyle(elements[i], null).zIndex || elements[i].currentStyle.zIndex;
        s = window.getComputedStyle(elements[i], null).position || elements[i].currentStyle.position;

        // Statically positioned elements are not affected by zIndex
        if (e && s !== "static") {
          found.push(parseInt(e, 10));
        }
    }

    return found.length ? Math.max.apply(null, found) : 0;
}



回答10:


Try my fiddle:

http://planitize.tumblr.com/post/23541747264/get-highest-z-index-with-descendants-included

This combines three advantages I haven't seen combined elsewhere:

  • Gets either the highest explicitly defined z-index (default) or the highest computed one.
  • Will look at all descendants of your selector, or all descendants of the document if none is supplied.
  • Will return either the value of the highest z, or the element that has the highest z.

One disadvantage: no cross-browser guarantees.




回答11:


If you are doing what I think you're doing, there is no need. Just do this:

$('div[id^=layer-]').css('z-index', 0);
$(this).css('z-index', 1000);


来源:https://stackoverflow.com/questions/5680770/how-to-find-the-highest-z-index-using-jquery

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