how to get different array on different index of single array - objective-c

Question

i have one array named invoiceInfo1 & i pass it to another array named allInfo.
But I want for different index of invoiceInfo ,the different array is created & pass it to allInfo.
Mycode is as follow:
for (int i=0; i<[userdata count]; i++) {
NSLog(#" userdata count :%d",[userdata count]);
invoiceInfo1 = [NSMutableArray arrayWithObjects:[[userdata objectAtIndex:i]valueForKey:#"fname"], [[userdata
objectAtIndex:i]valueForKey:#"name"],
[[userdata objectAtIndex:i]valueForKey:#"address"],
[[userdata objectAtIndex:i]valueForKey:#"city"], nil];
NSLog(#" info1 is:%#",invoiceInfo1); // invoiceInfo get overwrite when loop execute
NSMutableArray* allInfo = [NSMutableArray arrayWithObjects:headers,invoiceInfo1 , nil];
// HERE I Want Generate New Array of different index of invoiceInfo1 & pass it to allInfo
}

Answer 1

Create a NSObject class with header and info array:
#interface AllInfo: NSObject
{
NSMutableArray *header;
NSMutableArray *info;
}
#property (nonatomic, retain) NSMutableArray *header;
#property (nonatomic, retain) NSMutableArray *info;
#end
#implementation AllInfo
#synthesize header;
#synthesize info;
#end
Then implement this code:
allInfo = [[NSMutableArray alloc] init];
for (int i=0; i<[userdata count]; i++) {
NSLog(#" userdata count :%d",[userdata count]);
NSMutableArray *invoiceInfo1 = [[NSMutableArray alloc] init];
[invoiceInfo1 addObject:[userdata objectAtIndex:i]valueForKey:#"fname"]];
[invoiceInfo1 addObject:[userdata objectAtIndex:i]valueForKey:#"name"]];
[invoiceInfo1 addObject:[userdata objectAtIndex:i]valueForKey:#"address"]];
[invoiceInfo1 addObject:[userdata objectAtIndex:i]valueForKey:#"city"]];
NSLog(#" info1 is:%#",invoiceInfo1); // invoiceInfo get overwrite when loop execute
AllInfo *newInfo = [[AllInfo alloc] init];
newInfo.header = headers;
newInfo.info = invoiceInfo1;
[allInfo addObject:newInfo];
// HERE I Want Generate New Array of different index of invoiceInfo1 & pass it to allInfo
}
Hope this will help.