Gin 路由添加流程
作者:Z.K
- 2022-12-06 广东
本文字数:1982 字
阅读完需:约 7 分钟
计算 absolutePath
业务 HandlerFunc 与中间件 HandlerFunc 组合
Engine methodTrees 添加路由
methodTrees 实际上是 methodTree 组成的森林,每个 methodTree 是一颗 HTTP 方法的树
Engine.addRoute
func (engine *Engine) addRoute(method, path string, handlers HandlersChain) {
assert1(path[0] == '/', "path must begin with '/'")
assert1(method != "", "HTTP method can not be empty")
assert1(len(handlers) > 0, "there must be at least one handler")
debugPrintRoute(method, path, handlers)
root := engine.trees.get(method)//是否有HTTP方法的路由树
if root == nil {//没有 构建根节点
root = new(node)
root.fullPath = "/"
engine.trees = append(engine.trees, methodTree{method: method, root: root})
}
root.addRoute(path, handlers)
// Update maxParams
if paramsCount := countParams(path); paramsCount > engine.maxParams {
engine.maxParams = paramsCount
}
if sectionsCount := countSections(path); sectionsCount > engine.maxSections {
engine.maxSections = sectionsCount
}
}
复制代码
node.addRoute
func (n *node) addRoute(path string, handlers HandlersChain) {
fullPath := path
n.priority++
// Empty tree
if len(n.path) == 0 && len(n.children) == 0 {
n.insertChild(path, fullPath, handlers)
n.nType = root
return
}
parentFullPathIndex := 0
walk:
for {
// Find the longest common prefix.
// This also implies that the common prefix contains no ':' or '*'
// since the existing key can't contain those chars.
i := longestCommonPrefix(path, n.path)
// Split edge
if i < len(n.path) {
child := node{
path: n.path[i:],
wildChild: n.wildChild,
indices: n.indices,
children: n.children,
handlers: n.handlers,
priority: n.priority - 1,
fullPath: n.fullPath,
}
n.children = []*node{&child}
// []byte for proper unicode char conversion, see #65
n.indices = bytesconv.BytesToString([]byte{n.path[i]})
n.path = path[:i]
n.handlers = nil
n.wildChild = false
n.fullPath = fullPath[:parentFullPathIndex+i]
}
// Make new node a child of this node
if i < len(path) {
path = path[i:]
c := path[0]
// '/' after param
if n.nType == param && c == '/' && len(n.children) == 1 {
parentFullPathIndex += len(n.path)
n = n.children[0]
n.priority++
continue walk
}
// Check if a child with the next path byte exists
for i, max := 0, len(n.indices); i < max; i++ {
if c == n.indices[i] {
parentFullPathIndex += len(n.path)
i = n.incrementChildPrio(i)
n = n.children[i]
continue walk
}
}
// Otherwise insert it
if c != ':' && c != '*' && n.nType != catchAll {
// []byte for proper unicode char conversion, see #65
n.indices += bytesconv.BytesToString([]byte{c})
child := &node{
fullPath: fullPath,
}
n.addChild(child)
n.incrementChildPrio(len(n.indices) - 1)
n = child
} else if n.wildChild {
// inserting a wildcard node, need to check if it conflicts with the existing wildcard
n = n.children[len(n.children)-1]
n.priority++
// Check if the wildcard matches
if len(path) >= len(n.path) && n.path == path[:len(n.path)] &&
// Adding a child to a catchAll is not possible
n.nType != catchAll &&
// Check for longer wildcard, e.g. :name and :names
(len(n.path) >= len(path) || path[len(n.path)] == '/') {
continue walk
}
// Wildcard conflict
pathSeg := path
if n.nType != catchAll {
pathSeg = strings.SplitN(pathSeg, "/", 2)[0]
}
prefix := fullPath[:strings.Index(fullPath, pathSeg)] + n.path
panic("'" + pathSeg +
"' in new path '" + fullPath +
"' conflicts with existing wildcard '" + n.path +
"' in existing prefix '" + prefix +
"'")
}
n.insertChild(path, fullPath, handlers)
return
}
// Otherwise add handle to current node
if n.handlers != nil {
panic("handlers are already registered for path '" + fullPath + "'")
}
n.handlers = handlers
n.fullPath = fullPath
return
}
}
复制代码
划线
评论
复制
发布于: 刚刚阅读数: 5
版权声明: 本文为 InfoQ 作者【Z.K】的原创文章。
原文链接:【http://xie.infoq.cn/article/3a8047a5b6b9348fd054c948f】。文章转载请联系作者。
Z.K
关注
还未添加个人签名 2020-07-09 加入
还未添加个人简介
评论