Trouble with PHP if else logic and variables

Question

So, I have a couple of variables in PHP, and they are generated from different places. $new_ip comes from the following command:

$new_ip = file_get_contents('http://www.ipaddresscheck.comlu.com/ip.php');

And the $old_ip comes from:

$sql = mysql_query('SELECT * FROM ip ORDER BY  id DESC LIMIT 1');
$row = mysql_fetch_array( $sql );
$old_ip = $row['current_ip'];

My problem is, I am getting incorrect results from:

if ($old = $new)
{
$different = '1';    
}
else {
$different = '2';  
}
echo $different;

I ALWAYS get 2, wether the IPs are the same or not; if I use '==' as the comparison, I ALWAYS get 1.

When I run the following code, I get the following output: var_dump($old_ip); var_dump($new_ip);

Output

string(15) "123.123.123.123" string(167) "184.6.216.163 "

Are the variables different types? If So, can I make them the same so I am only comparing the IP and not the type? If the IPs are the same I should get '1' and if they are different, I should get '2', right?

You are getting string(167) on var_dump($new_ip)? Are you sure there is nothing else in that string as 184.6.216.163 is only 13. — Sean
– Sean, Commented Jun 28, 2013 at 5:07

user1864610 · Accepted Answer · 2013-06-28 05:06:18Z

4

Your problem is lurking here:

string(15) "123.123.123.123" string(167) "184.6.216.163 "
                                    ^^^

The address retrieved from your database is coming along with 154 unwanted characters. trim() both strings and then do the comparison.

Oh - and you should really shorten the field that stores the IP address. 15 characters should do.

answered Jun 28, 2013 at 5:06

user1864610

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2 Comments

user1801540 Over a year ago

The number on the left is coming from the database, the one on the right, is from the $new_ip = file_get_contents('ipaddresscheck.comlu.com/ip.php');

user1864610 Over a year ago

@JustinMarmorato Sorry - my mistake. I lost track of where the values were coming from in the midst of a phone call. Nevertheless, the value I highlighted is still way too long. Take a look at the code that creates it.

Rohan Kumar · Accepted Answer · 2013-06-28 05:09:19Z

2

Code

if ($old = $new) // you are using assignment operator here
{
   $different = '1';    
}

Should be

if ($old == $new) //Equal to operator
{
   $different = '1';    
}

Also confirm the variable names are $old and $new or $old_ip and $new_ip

edited Jun 28, 2013 at 5:09

answered Jun 28, 2013 at 5:02

Rohan Kumar

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2 Comments

i Code 4 Food Over a year ago

Quoted from the OP: if I use '==' as the comparison, I ALWAYS get 1.. So there is more to it.

Jack Daniel's Over a year ago

it should be $old_ip == $new_ip... you dont have $old variable in the code provided its $old_ip and $new_ip

Fasil kk · Accepted Answer · 2013-06-28 05:19:51Z

1

Assignment ( = ): Assigns the value on the right to the variable on the left
Equality ( == ): Checks if the left and right values are equal

Identical ( === ): Checks if the left and right values are equal AND identical (same variable type)

Example:

$a = 1; // Sets the value of $a as the integer 1
$b = TRUE; // Sets the value of $b to the boolean TRUE
if ($a == $b){
echo 'a is equal to b.';
}

if ($a === $b){ // compare VALUE and data type: if $a(integer) === $b(boolean)
echo 'a is identical and equal to b.';
}

if ($a = $b){ 
echo '$b value to variable $a';
}

In your code i noticed a problem. A space is there in second IP address. so

string(15) "123.123.123.123" string(167) "184.6.216.163 "
                                                       ^

Trim both IP address.
If not work, then convert it to other datatype and compare.(Just for debugging purpose. :))

edited Jun 28, 2013 at 5:19

answered Jun 28, 2013 at 5:07

Fasil kk

2,19717 silver badges27 bronze badges

4 Comments

user1801540 Over a year ago

how do I strip the whitespace? Ive tried $old_ip3 = str_replace(' ', '', $old_ip2); $new_ip3 = str_replace(' ', '', $new_ip2);

Fasil kk Over a year ago

Build-in function are there to trim PHP string. $old_ip2 = trim($old_ip);

user1801540 Over a year ago

Ive tried trim and str_replace. I can't seem to get the extra hundred something characters out of that variable. Is there a way I cam limit the size of the variable? That would be a start, then I would only have to figure out how to deal with the space when the IP is not 3.3.3.3, say 2.3.1.3.

Daniel P Over a year ago

You could use regular expression to strip your IP from the rest. Use preg_match() to find a match for a IP address and return the match, or use preg_replace() and replace all unwanted characters.

aman · Accepted Answer · 2013-06-28 05:13:35Z

0

OLD :

if ($old = $new){
   $different = '1';    
}

New Correct :

if ($old == $new){
   $different = '1';    
}

you are using (=) assignment operation in the place of comparison operation.

answered Jun 28, 2013 at 5:13

aman

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1 Comment

user1801540 Over a year ago

I actually did that before I posted the question. The question was a typo. But regardless, it still does not work.

Daniel P · Accepted Answer · 2013-06-28 07:26:07Z

As far as I can see the behaviour is normal because you are using $old and $new both are unset.

$new_ip = file_get_contents('http://www.ipaddresscheck.comlu.com/ip.php');

$sql = mysql_query('SELECT * FROM ip ORDER BY  id DESC LIMIT 1');
$row = mysql_fetch_array( $sql );
$old_ip = $row['current_ip'];

The next code will always give 2 because $new is unset. What you are doing here is assigning $old the value of $new and since $new is unset $old = $new will always be false.

if ($old = $new)
{
    $different = '1';    
} else {
    $different = '2';  
}
echo $different;

In the next example you are actually comparing $old and $new. Since both are unset the expression $old == $new will always be true.

if ($old == $new)
{
    $different = '1';    
} else {
    $different = '2';  
}
echo $different;

What you need to do is using the right vars and trim them.

// get new ip and trim white spaces
$new_ip = trim(file_get_contents('http://www.ipaddresscheck.comlu.com/ip.php'));
// get the old ip
$sql = mysql_query('SELECT * FROM ip ORDER BY  id DESC LIMIT 1');
$row = mysql_fetch_array( $sql );
$old_ip = $row['current_ip'];

if ($old_ip == $new_ip)
{
    $different = '1';    
} else {
    $different = '2';  
}
echo $different;

I am not trimming white spaces on $old_ip because I guess that you will be using $new_ip to update your database and since $new_ip is already trimmed we can be sure that $old_ip has no white spaces.

You should always have php warnings enabled when coding! It would have popped you a warning because you used unset vars.

PS : note that this code will only work as expected if file_get_contents() returns a IP with white spaces, the output could be different (error output, not able to detect ip correctly, ... etc)

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