what is the exact meaning of "message" in this line on iphone

Question

NSDictionary *story = [stories objectAtIndex: indexPath.row];
cell.text=[NSString stringwithFormat:[story objectForKey@"message];

i dont knw what exaclty "message " contains (what is the meaning of objectForKey@"message")

EDIT CODE

NSString *key =[appDelegate.books objectAtIndex:indexPath.row];
    //dict y=@"Name";

    NSArray *nameSection = [dict objectForKey:key];


    NSDictionary *story = [nameSection objectAtIndex: indexPath.row];
    cell.text=[NSString stringwithFormat:[story objectForKey:key]];

               NSLog(@"Value Of message: %@", [dict objectForKey:key]);

why my code crashes

Answer 1

@"message" is a key for a value stored in the NSDictionary object. The first line declares an NSDictionary named story that appears to come from an array.

If you want to find what value is stored for the key:@"message", consider using:

NSLog(@"Value Of message: %@", cell.text);

Run and check the console to see the output. (SHIFT + COMMAND + Y) in XCode will bring up the console, if that's what you are using. If you are unfamiliar with NSArrays/NSDictionaries, give Apple's Documentation a look.

I'm just guessing at all of this since that is a very limited sample of code. Try submit more code when you ask a question so that the viewers can get a better idea of your questions.

Answer 2

Hi,

That is an example of key-value coding, and a lot of information is available on the Apple dev site if you're interested:

http://developer.apple.com/mac/library/documentation/Cocoa/Conceptual/KeyValueCoding/KeyValueCoding.html

Answer 3

If you are more familiar with Java or C# the code is equivalent to something like this:

// Assuming stories is declared as: List<Dictionary<string, string> stories;

Dictionary<string, string> story = stories[indexPath.row];
cell.Text = String.Format(story["message"]);

In Smalltalk-style (and therefore Objective-C too) Object Oriented programming, methods are more like messages to other objects. So a good Objective-C method name should read like an English sentence (Subject-Verb-Object). Because of this working with dictionaries (hash tables) looks like this:

[myDictionary setObject:@"Value" forKey:@"someKey"];
[myDictionary objectForKey:@"someKey"]; // == @"Value"

In Java it would be:

myDictionary.put("someKey", "Value");
myDictionary.get("someKey"); // == "Value"

Notice how the key ("someKey") was the first argument in the Java example. In Objective-C you name your arguments with the method name, hence setObject: forKey:. Also notice that in Objective-C strings start with an @ symbol. That's because Objective-C strings are different from regular C strings. When using Objective-C you almost always use Objective-C's @ strings.

In C# there is a special syntax for Dictionaries so it becomes:

myDictionary["someKey"] = "Value";
myDictionary["someKey"]; // == "Value"

One important problem that you might encounter if you're new is the problem of native types.

In Java to add an int to a Dictionary you used to have to do:

myDictionary.put("someKey", new Integer(10));

Because the primitive types (int, char/short, byte, boolean) aren't real Objects. Objective-C has this problem too. So if you want to put an int into a dictionary you must use NSNumber like so:

[myDictionary setObject:[NSNumber numberForInt:10] 
                 forKey:@"someKey"];

And you pull out the integer like so:

NSNumber *number = [myDictionary objectForKey:@"someKey"];
[number intValue]; // == 10

EDIT:

Your code might be crashing if you have a '%' character in your string, since stringWithFormat is just like NSLog in that it takes many arguments. So if story["message"] is "Hello" then it'll work fine without extra arguments but if it's "Hello %@" you need to add one argument to stringWithFormat.

NSString *message = @"Hello %@";
NSMutableDictionary *dict = [NSMutableDictionary dictionary];
[dict setObject:message forKey:@"message"];

NSString *output = [NSString stringWithFormat:[dict objectForKey:@"message"], @"World!"];
// output is now @"Hello World!".