我怎么能轻松地做到这一点,同时使我的inNode仍然与传递它的函数相同呢?private void openNodeWindow(TreeNode inNode) // Find the top level window type this is in.TreeNode curNode = inNode;
// As long as therecurNode = curNod
我的密码查询是WITH path,[i in nodes(path) where 'InNode' in labels(i)] as InNodes
WHERE ALL (i IN InNodes WHERE i.Status in现在,我想检查这个InNode是否与属性检查有另一个Out关系。如果我在上面的代码中添加了下面的
std::pair<const long unsigned int, node*> >](((const long unsigned int&)((const long unsigned int*)(& inNode0; i < inID.size(); i++)}
nodeMap[inNode->getID()] = n