Searching an ArrayList for a string and returning another associated string

Question

private ArrayList<SocialMediaAccount> socialMediaAccounts;

public void addSocialMediaAccount(String userID, String websiteName, String websiteURL, int activityLevel) 
{
    SocialMediaAccount object = new SocialMediaAccount(userID, websiteName, websiteURL, activityLevel);
    socialMediaAccounts.add(object);
}

I have this ArrayList and I need to search for a particular websiteName and return the userID associated with the object. Getting confused a lot and would like some help on getting started with this problem. Thanks!

Hi refer to the comment I left on your solution.

hK-
– hK-

2017-04-02 08:23:11 +00:00
Commented Apr 2, 2017 at 8:23 — hK-
– hK-, Commented Apr 2, 2017 at 8:23

Faraz · Accepted Answer · 2017-04-02 03:49:18Z

2

You can go through the arraylist and check if the websiteName matches the websiteName that is stored in the list like this:

for(int i = 0; i < socialMediaAccounts.size(); i++)
    if(socialMediaAccounts.get(i).getWebSiteName().equals(the_website_you_arelooking_for){
    return socialMediaAccounts.get(i).getUserId
    }
}

edited Apr 2, 2017 at 3:49

answered Apr 2, 2017 at 3:38

Faraz

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4 Comments

Kevin Anderson Over a year ago

if needs an extra right paren, but overall premise seems sound.

hK- Over a year ago

Hi, the header code I have to use is public String getSocialMediaID(String websiteName)

hK- Over a year ago

Hi no, I'm not too sure how to use that header with the code you provided or with Rahul's solution.

hK- Over a year ago

@FarazDurrani Any help is appreciated

Kevin Anderson · Accepted Answer · 2017-04-02 03:53:38Z

1

If you have a lot of items in your list and you're doing a lot of searches, a HashMap using the the website name as the key will be much faster.

answered Apr 2, 2017 at 3:53

Kevin Anderson

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Comments

Rajith Pemabandu · Accepted Answer · 2017-04-02 03:55:54Z

1

I suppose this method will solve your question. Hope you have getters inside your socialMediaAccount class.

public String getuserID(ArrayList<SocialMediaAccount> socialMediaAccounts,String websiteName){
 for(SocialMediaAccount s:socialMediaAccounts){  
    if( s.getWebsiteName().equalsIgnoreCase( websiteName)){
        return s.getUserID;
    } 
} 
return "no user found";
}

edited Apr 2, 2017 at 3:55

answered Apr 2, 2017 at 3:38

Rajith Pemabandu

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4 Comments

Faraz Over a year ago

Rajith I think this answer is not correct. Because if the first condition fail, it will just return no user found and get out of the loop. don't you think?

Rajith Pemabandu Over a year ago

@Faraz you are correct.. I have modified the code.

Kevin Anderson Over a year ago

{} at end of if is suspect; shouldn't it be just {?

Rajith Pemabandu Over a year ago

@KevinAnderson my bad. I fixed.

Rahul Chowdhury · Accepted Answer · 2017-04-02 04:03:20Z

1

This should work.

//This method will return the userID associated with the given target websiteName.
//Insert it where you need it.
public String search(String targetWebsiteName){
    //loop through each account in your list.
    For(SocialMediaAccount acc: socialMediaAccounts){
      SocialMediaAccount tempObject = acc; 
       //check for websiteName
       if(tempObject.getWebsiteName().equals(targetWebsiteName)) return tempObject.getUserID();
    }
return null;
}


//Add these methods to your SocialMediaAccount class
class SocialMediaAccount{

       //Getters for object variables
       String getWebsiteName(){
            return websiteName;
       }

       String getUserID(){
           return userID;
       }
}

edited Apr 2, 2017 at 4:03

answered Apr 2, 2017 at 3:53

Rahul Chowdhury

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Comments

Fady Saad · Accepted Answer · 2017-04-02 03:36:07Z

0

I think what you will want    

for (SocialMediaAccount socialMediaAccount: socialMediaAccounts) {
            if (socialMediaAccount.getwebsiteName() == your_website_search_name) {
                return socialMediaAccount.getId();
            }
        }
        return null;

answered Apr 2, 2017 at 3:36

Fady Saad

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1 Comment

Kevin Anderson Over a year ago

== won't do it; you need to use socialMediaAccount.getWebSiteName().equals(your_website_search_name)

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