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in patchKeyedChildren fix unKeydNodes #105

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in patchKeyedChildren fix unKeydNodes #105

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19 changes: 11 additions & 8 deletions packages/runtime-core/src/renderer.ts
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Original file line number Diff line number Diff line change
Expand Up @@ -19,7 +19,7 @@ export function createRenderer(options) {
} = options;

const render = (vnode, container) => {
console.log("调用 patch")
console.log("调用 patch");
patch(null, vnode, container);
};

Expand Down Expand Up @@ -165,12 +165,11 @@ export function createRenderer(options) {
mountChildren(c2, container);
} else {
// array diff array
// 如果之前是 array_children
// 现在还是 array_children 的话
// 那么我们就需要对比两个 children 啦
// 如果之前是 array_children
// 现在还是 array_children 的话
// 那么我们就需要对比两个 children 啦
patchKeyedChildren(c1, c2, container, parentComponent, anchor);
}

}
}

Expand Down Expand Up @@ -240,7 +239,7 @@ export function createRenderer(options) {
}
} else if (i > e2 && i <= e1) {
// 这种情况的话说明新节点的数量是小于旧节点的数量的
// 那么我们就需要把多余的
// 那么我们就需要把多余的节点给删除掉
while (i <= e1) {
console.log(`需要删除当前的 vnode: ${c1[i].key}`);
hostRemove(c1[i].el);
Expand Down Expand Up @@ -288,14 +287,18 @@ export function createRenderer(options) {

let newIndex;
if (prevChild.key != null) {
// 这里就可以通过key快速的查找了， 看看在新的里面这个节点存在不存在
// 这里就可以通过key快速的查找了，看看在新的里面这个节点存在不存在
// 时间复杂度O(1)
newIndex = keyToNewIndexMap.get(prevChild.key);
} else {
// 如果没key 的话，那么只能是遍历所有的新节点来确定当前节点存在不存在了
// 时间复杂度O(n)
for (let j = s2; j <= e2; j++) {
if (isSameVNodeType(prevChild, c2[j])) {
// 只有在这个索引位置未被赋值时(为0时,这样才会依次记录下所有没有key的节点,否则会一直覆盖记录同一个节点),并且是同类型的节点才会进行记录
if (
newIndexToOldIndexMap[j - s2] === 0 &&
isSameVNodeType(prevChild, c2[j])
) {
newIndex = j;
break;
}
Expand Down