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避免 menuCommand 更新在 Tab 移除后触发 #828

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Merged
CodFrm merged 2 commits into from
Oct 10, 2025
Merged

避免 menuCommand 更新在 Tab 移除后触发 #828

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避免 menuCommand 更新在 Tab 移除后触发
cyfung1031 Oct 10, 2025
5872c7b
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cyfung1031 Oct 10, 2025
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7 changes: 6 additions & 1 deletion src/app/service/service_worker/popup.ts
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Original file line number Diff line number Diff line change
Expand Up @@ -179,7 +179,6 @@ export class PopupService {

// 防止并发导致频繁更新菜单，将注册菜单的请求集中在一个队列中处理
updateMenuCommands = new Map<number, ((TScriptMenuRegister | TScriptMenuUnregister) & { registerType: number })[]>();
isUpdateMenuDirty = false;

// 此函数必须是同步执行的，避免updateMenuCommands并发问题
updateMenuCommand(tabId: number, data: ScriptMenu[]): string[] {
Expand Down Expand Up @@ -579,6 +578,12 @@ export class PopupService {
}
runCountMap.delete(tabId);
scriptCountMap.delete(tabId);
const list = this.updateMenuCommands.get(tabId);
if (list) {
// 避免 menuCommand 更新在 Tab 移除后触发
list.length = 0;
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this.updateMenuCommands.delete(tabId);
}
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@CodFrm CodFrm Oct 10, 2025 •
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我理解 this.updateMenuCommand 在 updateMenuCommand 最后面肯定会删除的，且updateMenuCommand是一个同步函数，这实际是一个无效操作

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@cyfung1031 cyfung1031 Oct 10, 2025

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呀。原来现在是 ScriptMenu 没删 （${CACHE_KEY_TAB_SCRIPT}${tabId}）
对。会跑。所以如果推了10个registerMenu, onRemoved 后才执行，那么就会变成在 script.menus 里残留

// 清理数据tab关闭需要释放的数据
this.txUpdateScriptMenu(tabId, async (scripts) => {
for (const { uuid } of scripts) {
Expand Down
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