CODEWITHSUNDEEP
php
Azeem
Azeem

Reputation: 2924

Undefined variable Error when try to access PHP class variable in function

I have run into a problem. My php class structure is as follows:

    class CustomerDao{
...
var $lastid;

  function insertUser($user)
  {
    ...
    $lastid = mysql_insert_id();
    return 0;
  }
      function getCustId()
  { 
    return $lastid; 
  }
    }

When i use this class, it let me access $lastid varibale in first function "insertUser", but it throws an error when i use $lastid in second function. I have no idea how to resolve this problem. Please guide.

Upvotes: 0

Views: 1817

Answers (6)

Quasdunk
Quasdunk

Reputation: 15230

What you want to do is this:

function insertUser($user) {
  ...
  $this->lastid = mysql_insert_id();
  return 0;
}

function getCustId() { 
  return $this->lastid; 
}

Note the this-keyword. Your first function works, because you assign a new (local!) variable $lastid within your insertUser() function - but it has nothing to do with the class property $lastid.

Upvotes: 2

Treffynnon
Treffynnon

Reputation: 21563

Your code sample should look this:

class CustomerDao{
...
var $lastid;

  function insertUser($user)
  {
    ...
    $this->lastid = mysql_insert_id();
    return 0;
  }
      function getCustId()
  { 
    return $this->lastid; 
  }
    }

You need to reference the class ($this) to access its $lastid property. So it should be $this->lastid;

Upvotes: 3

Dau
Dau

Reputation: 8858

to use a class variable inside the class use the $this keyword

so to use $lastid variable inside class use $this->lastid

Upvotes: 2

shanethehat
shanethehat

Reputation: 15580

In your first function you are creating a new variable called $lastid which exists only within the scope of the function. In the second function this fails because there is no $lastid variable declared within this function.

To access a class member you use the notation $this->lastid.

class CustomerDao {
    ...
    var $lastid;

    function insertUser($user)
    {
        ...
        $this->lastid = mysql_insert_id();
        return 0;
    }

    function getCustId()
    { 
        return $this->lastid; 
    }
}

Upvotes: 4

Pekka
Pekka

Reputation: 449783

If you want to change an object property, you want the this keyword:

$this->lastid = mysql_insert_id();

Reference: PHP Manual: Classes and objects

Upvotes: 5

fivedigit
fivedigit

Reputation: 18692

You're trying to access a class variable, which is done like this instead:

function getCustId() { 
    return $this->lastid; 
}

Upvotes: 7

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