蘋果禁用 UIWebView

We identified one or more issues with a recent delivery for your app, XXXXXX. Please correct the following issues, then upload again.

ITMS-90809: Deprecated API Usage?- New apps that use UIWebView are no longer accepted. Instead, use WKWebView for improved security and reliability. Learn more

5月開始,蘋果已經(jīng)嚴格??使用UIWebView,只要是新賬號&新包會出現(xiàn)以上郵件信息,開發(fā)者該如何檢測第三方SDK中是否含有UIWebView呢?希望可以幫助最近要上架的小伙伴~~~

find . -type f | grep -e ".a" -e ".framework" | xargs grep -s UIWebView


./Pods/SensorsAnalyticsSDK/SensorsAnalyticsSDK/SensorsAnalyticsSDK.m:? ? ? ? UIWebView *webView = [[UIWebView alloc] initWithFrame:CGRectZero];

./Pods/SensorsAnalyticsSDK/SensorsAnalyticsSDK/SensorsAnalyticsSDK.m:? ? ? ? NSAssert(![webView isKindOfClass:NSClassFromString(@"UIWebView")], @"當(dāng)前集成方式已禁用 UIWebView!?");

./Pods/SensorsAnalyticsSDK/SensorsAnalyticsSDK/SensorsAnalyticsSDK.m:? ? ? ? if ([webView isKindOfClass:[UIWebView class]]) {//UIWebView

./Pods/SensorsAnalyticsSDK/SensorsAnalyticsSDK/SensorsAnalyticsSDK.m:? ? ? ? ? ? SALogDebug(@"showUpWebView: UIWebView");

./Pods/SensorsAnalyticsSDK/SensorsAnalyticsSDK/SensorsAnalyticsSDK.m:? ? ? ? ? ? SALogDebug(@"showUpWebView: not UIWebView or WKWebView");

./Pods/SensorsAnalyticsSDK/SensorsAnalyticsSDK/SensorsAnalyticsSDK.h: * @param webView 當(dāng)前 WebView,支持 UIWebView 和 WKWebView

./Pods/SensorsAnalyticsSDK/SensorsAnalyticsSDK/SensorsAnalyticsSDK.h: * @param webView 當(dāng)前 WebView,支持 UIWebView 和 WKWebView

./Pods/SensorsAnalyticsSDK/SensorsAnalyticsSDK/VisualizedAutoTrack/SAVisualizedAutoTrackObjectSerializer.m:? ? ? ? [NSStringFromClass(object.class) isEqualToString:@"UIWebView"] ||

./Pods/SensorsAnalyticsSDK/SensorsAnalyticsSDK/VisualizedAutoTrack/SAVisualizedAutoTrackObjectSerializer.m:? ? ? ? [object isKindOfClass:UIWebView.class] ||

最后編輯于
?著作權(quán)歸作者所有,轉(zhuǎn)載或內(nèi)容合作請聯(lián)系作者
【社區(qū)內(nèi)容提示】社區(qū)部分內(nèi)容疑似由AI輔助生成,瀏覽時請結(jié)合常識與多方信息審慎甄別。
平臺聲明:文章內(nèi)容(如有圖片或視頻亦包括在內(nèi))由作者上傳并發(fā)布,文章內(nèi)容僅代表作者本人觀點,簡書系信息發(fā)布平臺,僅提供信息存儲服務(wù)。

友情鏈接更多精彩內(nèi)容