Combining multiple strings in Cocoa - ios

Question

I want to be able to combine each index so that I get #"Biology Teacher A BK 1", but so far I have been unsuccessful. This is what I have so far, but I do not know where to go from here.
#interface ListTableViewController () <UISearchDisplayDelegate>
#property (strong, nonatomic) NSArray *className;
#property (strong, nonatomic) NSArray *teacherName;
#property (strong, nonatomic) NSArray *blockNumber;
#end
#implementation ListTableViewController
- (void)viewDidLoad {
[super viewDidLoad];
self.className = [NSArray arrayWithObjects:
#"Biology",
#"English III",
#"Chemistry",
#"Algebra II",
#"Morality", nil];
self.teacherName = [NSArray arrayWithObjects:
#"Teacher A",
#"Teacher B",
#"Teacher C",
#"Teacher D",
#"Teacher E", nil];
self.blockNumber = [NSArray arrayWithObjects:
#"BK 1",
#"BK 3",
#"BK 6",
#"BK 2",
#"BK 1", nil];
}

Answer 1

It Will work:
for (int i = 0 ; i< self.className.count; i++)
{
NSString *temStr = [NSString stringWithFormat:#"%# %# %#",[self.className objectAtIndex:i] ,[self.teacherName objectAtIndex:i],[self.blockNumber objectAtIndex:i] ];
NSLog("%#", tempStr);
}

Answer 2

Try this...
int total = self.className.count;
NSMutableArray *combinedName = [NSMutableArray array];
if (total == self.teacherName.count && total == self.blockNumber.count)
{
for(int i=0;i< total;i++)
{
NSString *str =[NSString stringWithFormat:#"%# %# %#", [self.className objectAtIndex:i],[self.teacherName objectAtIndex:i],[self. blockNumber objectAtIndex:i]];
[combinedName addObject:str];
}
}
else
NSLog(#"Cann't combine");

Answer 3

Try this
//Assuming three array are in same length
NSMutableArray *combineArray=[[NSMutableArray alloc] init];
for(int i=0; i<[[self className] count]; i++)
{
[combineArray addObject:[NSString stringWithFormat:#"%# %# %#", [[self className] objectAtIndex:i],[[self teacherName] objectAtIndex:i], [[self blockNumber] objectAtIndex:i]];
}
NSLog(#"%#", combineArray); //here is your output.

Answer 4

You can try this:
NSMutableArray *combinedArray = [[NSMutableArray alloc]init];
for (int i = 0; i < [self.className count]; i++)
{
NSString *combinedString = [NSString stringWithFormat:#"%# %# %#",[self.className objectAtIndex:i],[self.teacherName objectAtIndex:i],[self. blockNumber objectAtIndex:i]];
[combinedArray addObject:combinedString];
}
NSLog(#"Combined array is :\n %#",combinedArray);

Answer 5

Try with following code:
for (int i = 0 ; i< self.className.count; i++)
{
NSString *temStr = [[NSString stringWithFormat:#"%# ", [self.className objectAtIndex:i]] stringByAppendingString:[NSString stringWithFormat:#"%# ", [self.className objectAtIndex:i]]]
NSLog("%#", [temStr stringByAppendingString:[self.blockNumber objectAtIndex:i]])
}

Answer 6

Something like this would work, fairly ugly though.
self.className = [NSArray arrayWithObjects:#"Biology",#"English III",#"Chemistry",#"Algebra II",#"Morality", nil];
self.teacherName = [NSArray arrayWithObjects:#"Teacher A",#"Teacher B",#"Teacher C",#"Teacher D",#"Teacher E", nil];
self.blockNumber = [NSArray arrayWithObjects:#"BK 1",#"BK 3",#"BK 6",#"BK 2",#"BK 1", nil];
NSMutableArray *combinedNames = [[NSMutableArray alloc] init];
if (([self.className count] == [self.teacherName count]) && [self.className count] == [self.blockNumber count]) {
for (int index = 0; index < [self.className count]; index++) {
[combinedNames addObject:[NSString stringWithFormat:#"%# %# %#", [self.className objectAtIndex:index], [self.teacherName objectAtIndex:index], [self.blockNumber objectAtIndex:index]]];
}
}
for (NSString *string in combinedNames) {
NSLog(#"%#", string);
}
and that outputs:
Biology Teacher A BK 1
English III Teacher B BK 3
Chemistry Teacher C BK 6
Algebra II Teacher D BK 2
Morality Teacher E BK 1
Update
Looks like this was posted already by others before I could finish getting it put together. I don't see that they verify that the array's are all of the same length though. You can use anyones answer; it might be wise to verify that all of the array's contain the same number of objects prior to trying to iterate through them.