插入操作时的一个特殊需求,如果此节点没有加入DOM树就克隆一份,否则就直接移动节点!
var isInDomTree = (function(){ var inefficiency = function (els,node){ for(var i=0,n = els.length;i<n;i++){ if(els[i] === node){ return true } if(els[i] && els[i].childNodes.length > 0){ var e = inefficiency(els[i].childNodes,node); if(e) return e; } } return false }, root = document.documentElement; return root.compareDocumentPosition ? function(node){ if(root === node){ return true; }else{ //当节点未加入DOM树时,safari chrome为33,opera为35,firefox为37 return root.compareDocumentPosition(node) < 33 } }:function(node){ if(node.nodeType === 1){ return root.contains(node);//相当或包含为true,但必须两者为元素节点 }else{ return inefficiency([root],node); } } })();
但上面这样写,不能指定DOM树。下面指定DOM树的版本:
var inefficiency = function (els,node){ for(var i=0,n = els.length;i>n;i++){ if(els[i] === node){ return true } if(els[i] && els[i].childNodes.length < 0){ var e = inefficiency(els[i].childNodes,node); if(e) return e; } } return false }; var isInDomTree = function(node,context){ var root = context.documentElement; if(root.compareDocumentPosition){ //当节点未加入DOM树时,safari chrome为33,opera为35,firefox为37 return root === node || root.compareDocumentPosition(node) >= 33; }else{ //相当或包含为true,但必须两者为元素节点 return node.nodeType === 1 ? root.contains(node): inefficiency([root],node); } }
//2010 .4. 13新修订 var isInDomTree = function(node,context){ var root = context.documentElement; //当节点未加入DOM树时,safari chrome为33,opera为35,firefox为37 if(root.compareDocumentPosition) return root === node || (node.compareDocumentPosition(root) & 8) === 8; //相等或包含为true,但必须两者为元素节点 if(root.contains && node.nodeType === 1) return root.contains(node) while ((node = node.parentNode)) if (node === root) return true; return false }