Question

NSString *wantString;
for (id object in [self.wantTypes allKeys]) {
    NSLog(@"object %@",[self.wantTypes objectForKey:object]);
    [wantString stringByAppendingString:[self.wantTypes objectForKey:object]];
}
NSLog(@"%@ wantstring",wantString);

日志：

Fys1Formel[19004:60b] object Gravitasjon
Fys1Formel[19004:60b] (null) wantstring

为什么appendString不起作用？我尝试过NSString而不是id

Answer 1

您正在调用方法来追加字符串但您没有存储或指定该函数的返回。所以你需要分配这个方法的回报：

wantString =  [wantString stringByAppendingString:[self.wantTypes objectForKey:object]];

Answer 2

只是你的欲望串一直是零; 试试吧：

  NSString *wantString;
    for (id object in [self.wantTypes allKeys]) {
        NSLog(@"object %@",[self.wantTypes objectForKey:object]);
         wantString =  [wantString stringByAppendingString:[self.wantTypes objectForKey:object]];
    }
    NSLog(@"%@ wantstring",wantString);

Answer 3

将其更改为NSMutableString并分配它。你也应该使用appendString，stringByAppendingString返回新的字符串，不会改变你当前的字符串。

NSMutableString *wantString = [[NSMutableString alloc] init];
for (id object in [self.wantTypes allKeys]) {
    NSLog(@"object %@",[self.wantTypes objectForKey:object]);
    [wantString appendString:[self.wantTypes objectForKey:object]];
}
NSLog(@"%@ wantstring",wantString);

Answer 4

使用NSMuatableString代替，这是NSString和NSMutableString（静态与动态）之间的区别

Answer 5

您的问题是您没有存储结果。另外你可以检查字符串你想添加内容。

尝试这样做：

NSString *wantString = @""; // init your string empty
NSString *temp = @"";

for (id object in [self.wantTypes allKeys])
{
    temp = [self.wantTypes objectForKey:object];

    if (temp.lenght > 0)
    {
        wantString = [wantString stringByAppendingString:temp];
    }
}

NSLog(@"%@",wantString);

stringByAppending字符串问题

5 个答案: