Checking if word is valid in array

Question

I am using code to check if a word is in my array, if it is I want it to submit it and I have the code for it. If it isn't I want it to pop up a screen. now this all works, the only thing is the screen pops up 2 times, because there are 2 words in my array. here is the code to explain it a little better.

NSArray *searchContacts = [NSArray arrayWithObjects:
                           @"CADEAU",
                           @"KADERZ",nil];
NSString *myContact = labelsText.text;

for (NSString *contact in searchContacts) {
    if ([contact isEqualToString:myContact]) {

this is where I put in my words, CADEAU & KADERZ in this case. When I put one of these words into labelsText.text it does exactly what I want. but for the else statement if the labels text.text word is not CADEAU or KADERZ, it pop ups a screen:

else {
 UIAlertView *alert = [[UIAlertView alloc]

This screen will pup up 2 times now, so i'll have to press dismiss 2 times, how would I fix this to just have to press dismiss one time no mather how many words are in the array?

Michael Robinson · Accepted Answer · 2012-05-10 22:02:30Z

5

It would be more efficient to use an NSSet, but even if you use an NSArray, you can simply call containsObject: instead of looping through the collection yourself.

if (![searchContacts containsObject:myContact]) {
   //show alert...
}

edited May 10, 2012 at 22:02

Michael Robinson

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answered May 10, 2012 at 22:00

omz

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4 Comments

Kevin Over a year ago

oke is there any difference? I need to put in 12.5k words so any important thing I need to worry?

Michael Robinson Over a year ago

If all you want to do is determine whether a name is present in your array, then this is a good way to do it.

omz Over a year ago

If really all you want to do is to check whether some name is on your list of valid names and the order of that list is not important, you should use an NSSet or NSMutableSet instead of an NSArray. It is much faster for membership testing.

abarnert Over a year ago

For a large collection, a set is definitely going to be faster.

Michael Robinson · Accepted Answer · 2012-05-10 21:58:55Z

3

Put a break; after the code showing your alert:

for (NSString *contact in searchContacts) {
    if ([contact isEqualToString:myContact]) {
        // do something
    } else { 
        // show screen
        break;
    }
}

This will 'break' out of the loop.

answered May 10, 2012 at 21:58

Michael Robinson

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1 Comment

abarnert Over a year ago

Hold on, are you sure this is what you want? If you put in KADERZ, it's going to show the screen and fail, which I don't think is what you intended. If you really want to do it this way, see Matt's answer below. But omz's answer (using NSSet) is probably the best.

Matt · Accepted Answer · 2012-05-10 22:04:17Z

0

I think you want something like this:

    BOOL contactFound = NO;
    for (NSString *contact in array)
    {
        if ([contact isEqualToString:myContact])
        {
            contactFound = YES;
            break;
        }
    }

    if (!contactFound)
        UIAlertView *alert = [[UIAlertView alloc]...

answered May 10, 2012 at 22:04

Matt

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Comments

Tobias · Accepted Answer · 2012-05-10 22:09:20Z

0

Use a break after your UIAlertView.

For example:

for (NSString *contact in searchContacts) {
        if ([contact isEqualToString:myContact]) {
            //do what you want to do      
        }
        else{
            UIAlertView *alert = [[UIAlertView alloc] init];
            [alert show];
            break; //leave for()
        }
    }

Or use that:

searchContacts enumerateObjectsUsingBlock:^(id obj, NSUInteger idx, BOOL *stop) {
    if ([contact isEqualToString:myContact]) {
        //do what you want to do      
    }
    else{
        UIAlertView *alert = [[UIAlertView alloc] init];
        [alert show];

        *stop = YES; //stop enumeration
    }
}

edited May 10, 2012 at 22:09

answered May 10, 2012 at 22:03

Tobias

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1 Comment

abarnert Over a year ago

The is the same as the accepted answer, and it's wrong for the same reason: It will only accept the first value in searchContacts.

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