Concatenation of String with Array elements in Objective-C

Question

Can someone one explain to me why this code doesn't work? I have no problem sending the elements of an array to NSLog but they don't seem to be appending to the string. Do I have to cast the array elements to a string?

    success:^(NSURLRequest *request, NSHTTPURLResponse *response, id JSON) {
        NSArray *dataarray=[JSON valueForKey:@"Data"];
        NSLog(@"Response: %@", [JSON valueForKeyPath:@"Status"]);
        NSString* output = [NSString stringWithFormat:@"response: %@",[JSON valueForKeyPath:@"Status"]];
        int x;
        for (x=0; x<[dataarray count]; x++) {
            NSLog(@"%d : %@",x, [dataarray objectAtIndex:x]);
            [output stringByAppendingFormat:@" %@ ",[dataarray objectAtIndex:x]];
        }

        //NSLog(@"%@", JSON);
        NSLog(@"%@", output);
        self.outPut2.text=output;    }

Ismael · Accepted Answer · 2012-08-30 15:09:04Z

3

The function

 [output stringByAppendingFormat:@" %@ ",[dataarray objectAtIndex:x]];

returns a NEW string, without modifying the original, and you are not storing it anywhere. You should go like this:

output = [output stringByAppendingFormat:@" %@ ",[dataarray objectAtIndex:x]];

answered Aug 30, 2012 at 15:09

Ismael

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3 Comments

Codeguy007 Over a year ago

THanks for the quick response. This makes perfect sense

jere Over a year ago

Although this does work, it's really not the best way to go. If you are dynamically adding content to a string you should really use NSMutableString. After all, that's what is what meant for.

Codeguy007 Over a year ago

I could use a mutable variable but I wanted to know what I was doing wrong. The output to the label will probably be removed before the final app is complete. It's just there to show that the json based message passing code works.

Sixten Otto · Accepted Answer · 2012-08-30 15:10:04Z

2

Your output variable is an immutable NSString. -stringByAppendingFormat: doesn't append the new string in place, it returns a new string value that is the concatenation of the two original strings. You need to assign that value back to output.

In the alternative, make output an NSMutableString, and then you can do the concatenation in place with -appendFormat:.

answered Aug 30, 2012 at 15:10

Sixten Otto

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1 Comment

Codeguy007 Over a year ago

Thanks that makes perfect sense.

jere · Accepted Answer · 2012-08-30 15:09:50Z

1

You output variable is set to be an inmutable string. So you can't directly add any content to it. You can create a new string using it's content and reassign it to itself, but you can't append new content.

You should try using NSMutableString and appendFormat or appendString

        NSMutableString* output = [NSMutableString stringWithFormat:@"response: %@",[JSON valueForKeyPath:@"Status"]];
        int x;
        for (x=0; x<[dataarray count]; x++) {
            NSLog(@"%d : %@",x, [dataarray objectAtIndex:x]);
            [output appendFormat:@" %@ ",[dataarray objectAtIndex:x]];
        }

answered Aug 30, 2012 at 15:09

jere

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