error:[NSNull length]: unrecognized selector sent to instance 0x198dd6690'
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最近做请求数据,解析json中url出现的问题,
error:
[NSNull length]: unrecognized selector sent to instance 0x198dd6690'
solution:
从json数据中拿到字符串,有时候可能是空,这时判断长度就会崩溃,控制台输出是
-[NSNull length]: unrecognized selector sent to instance 0x6c8f6b0
那么就需要判断所得数据是不是NSNull才行,然而直接判断会出现如下警告:
以下是苹果官方文档的代码 :
id aValue = [arrayWithNull objectAtIndex:0]; if (aValue == nil) { NSLog(@"equals nil"); } else if (aValue == [NSNull null]) { NSLog(@"equals NSNull instance"); if ([aValue isEqual:nil]) { NSLog(@"isEqual:nil"); } }
那么判断时就可以:
if (eJid isEqual:[NSNull null])
如此,就解决了数据位null的判断 ;
另外,还有一种方法:
if (![name isKindOfClass:[NSNull class]]) {
}
也可以
0 0
- error:[NSNull length]: unrecognized selector sent to instance 0x198dd6690'
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