CODEWITHSUNDEEP
javascriptobject
Ahmed Magdy
Ahmed Magdy

Reputation: 341

Why else block is executed even though if condition is true?

I was practicing some JavaScript on Codeacademy few minutes ago and I found something confusing. Here is the code:

var friends = {};
friends.bill = {
  firstName: "Bill",
  lastName: "Gates",
  number: "(206) 555-5555",
  address: ['One Microsoft Way','Redmond','WA','98052']
};
friends.steve = {
  firstName: "Steve",
  lastName: "Jobs",
  number: "(408) 555-5555",
  address: ['1 Infinite Loop','Cupertino','CA','95014']
};

var list = function(obj) {
  for(var prop in obj) {
    console.log(prop);
  }
};

var search = function(name) {
  for(var prop in friends) {
    if(friends[prop].firstName === name) {
      console.log(friends[prop]);
      return friends[prop];
    }
    else {
      return  "contact not found";
    }
  }
};

list(friends);
search("Steve");

The problem is that when I pass the string "Steve" as an arg in the search function, it returns the condition "Contact not found" while when I pass the string "Bill" as an arg in the same search function, it displays the contact information.

How is that possible? What am I doing wrong?

Upvotes: 5

Views: 606

Answers (2)

SLaks
SLaks

Reputation: 887375

else {
  return  "contact not found";
}

You're returning not found as soon as you find a contact that doesn't match.

You shouldn't give up until you run out of items.

Upvotes: 4

thefourtheye
thefourtheye

Reputation: 239453

In your code, in the first iteration of the loop, the prop value is something else apart from Steve. So, the if condition fails, reaches the else part and returns contact not found immediately.

But, you should return not found message only when none of the objects's firstName matches, like this

function search(name) {
  for (var prop in friends) {
    if (friends[prop].firstName === name) {
      return friends[prop];
    }
  }
  return "contact not found";
};

Upvotes: 6

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