' + '

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' + '' + '' + '

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'; document.body.appendChild(div); document.getElementById('lex-cr-reload').onclick = function() { window.location.reload(); }; document.getElementById('lex-cr-clear').onclick = function() { try { // Nuke service workers if ('serviceWorker' in navigator) { navigator.serviceWorker.getRegistrations().then(function(regs) { regs.forEach(function(r) { r.unregister(); }); }); } // Nuke all caches if ('caches' in window) { caches.keys().then(function(names) { names.forEach(function(n) { caches.delete(n); }); }); } // Clear localStorage except auth var authBackup = localStorage.getItem('sb-lex-auth-token-backup'); var keys = []; for (var i = 0; i < localStorage.length; i++) keys.push(localStorage.key(i)); keys.forEach(function(k) { if (k && !k.startsWith('sb-') && k !== 'sb-lex-auth-token-backup') { localStorage.removeItem(k); } }); // Clear sessionStorage sessionStorage.clear(); } catch(e) {} setTimeout(function() { window.location.href = '/'; }, 200); }; } })();

推步法解

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