检测 Tab 重复事件

Question

我正在开发一个 Chrome 扩展；

我需要检测选项卡何时重复，我正在寻找选项卡检测事件，例如onduplicate.addListener()？

Chrome 扩展 API 可以实现这一点吗？

Answer 1

这是最接近的实现：

const newTabsId = new Set();

// Observe all new tabs with opener ID
chrome.tabs.onCreated.addListener(tab => {
    if(tab.openerTabId) {
        newTabsId.add(tab.id);
    }
});

// Waiting for a new tab completeness
chrome.tabs.onUpdated.addListener((tabId, changes, tab) => {
    if(newTabsId.has(tabId) && changes.status === 'complete') { 
        if(!tab.openerTabId) {
            return;
        }
        // Retrieve opener (original) tab
        getTabById(tab.openerTabId)
            .then(originalTab => {
                if(
                    originalTab.url === tab.url &&          // original and new tab should have same URL
                    originalTab.index + 1 === tab.index &&  // new tab should have next index
                    tab.active && tab.selected              // new tab should be active and selected
                                                            // also we may compare scroll from top, but for that we need to use content-script
                ) {
                    console.log('Duplicate:', tab);
                }
            });
        // Remove this tab from observable list
        newTabsId.delete(tabId);
    }
});

// Syntax sugar
function getTabById(id) {
    return new Promise((resolve, reject) => {
        chrome.tabs.get(id, resolve);
    });
}

// Cleanup memory: remove from observables if tab has been closed
chrome.tabs.onRemoved.addListener(tabId => {
    newTabsId.delete(tabId);
});

编辑 1：但是，现在没有明确的解决方案来检测真正的重复项