url = [NSString stringWithFormat:str.length >0?str:@"https://wa.me/8526210980?text=企業銀行同業"];
NSLog(@"url = %@",url);
url = (null)
url不能赋值成功
解决:因为包含了url不能识别的字
url = [NSString stringWithFormat:str.length >0?str:@"https://wa.me/8526210980?text=企業銀行同業"];
NSLog(@"url = %@",url);
url = (null)
url不能赋值成功
解决:因为包含了url不能识别的字
本文标题:url打印为空
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