Question

我想以下列格式创建一个字符串：id[]=%@&stringdata[]=%@&id[]=%@&stringdata[]=%@&id[]=%@&stringdata[]=%@&等在for循环中我获取id和stringdata，但是当我打印出testString时会说（null），但是当我打印出来时Ident和Stringdata单独（前两个NSLog）id确实打印出来，stringdata也显示（null）因为那时字符串是空的，所以这是正常的。

for(NSUInteger i = 0; i<self.childViewControllers.count; i++) {

        NSLog(@"Ident: %@",((myViewController*)[self.childViewControllers objectAtIndex:i]).ident);
        NSLog(@"Stringdata: %@",((myViewController*)[self.childViewControllers objectAtIndex:i]).getQAnswer);

        testString = [testString stringByAppendingString:([NSString stringWithFormat: @"id[]=%@&stringdata[]=%@&",((myViewController*)[self.childViewControllers objectAtIndex:i]).ident ,((myViewController*)[self.childViewControllers objectAtIndex:i]).getQAnswer])];

        NSLog(@"PostString : %@",testString);
}

Answer 1

这意味着当您进入此循环时testString为nil。因此，每次调用stringByAppendingString:时，都会将其传递给nil对象，该对象不执行任何操作。在循环结束时，testString仍为nil.

Answer 2

NSMutableString *mString = [NSMutableString string];
for(NSUInteger i = 0; i<self.childViewControllers.count; i++)
{
    [mString appendFormat:@"%@id[]=%@",(i>0)?@"&":@"",((myViewController*)[self.childViewControllers objectAtIndex:i]).ident];
    [mString appendFormat:@"&stringdata[]=%@",((myViewController*)[self.childViewControllers objectAtIndex:i]).getQAnswer];
}
NSLog(@"PostString : %@",mString);

希望这会有所帮助

使用测试字符串输出：

2013-04-04 23：46：54.619 PagingScrollView [93543：c07] PostString： ID [] = TESTI＆安培; StringData是[] = TESTQ和ID [] = TESTI＆安培; StringData是[] = TESTQ和ID [] = TESTI＆安培; StringData是[] = TESTQ和ID [] = TESTI＆安培; StringData是[] = TESTQ和ID [ ] = TESTI＆安培; StringData是[] = TESTQ和ID [] = TESTI＆安培; StringData是[] = TESTQ和ID [] = TESTI＆安培; StringData是[] = TESTQ和ID [] = TESTI＆安培; StringData是[] = TESTQ和ID [] = TESTI＆安培; StringData是[] = TESTQ和ID [] = TESTI＆安培; StringData是[] = TESTQ

Answer 3

您在testString上添加了该格式。由于未初始化，因此nil和nil的任何消息都会返回nil。

您应该使用：

testString = [NSString stringWithFormat: @"id[]=%@&stringdata[]=%@&",((myViewController*)[self.childViewControllers objectAtIndex:i]).ident ,((myViewController*)[self.childViewControllers objectAtIndex:i]).getQAnswer];

打印出连接的字符串显示（null）

3 个答案: