Difference between numeric and string value in PHP - php

I am getting a difference when comparing two string which are 0 and '0' in PHP. Can I do anything to make them be compared equally in an if action?
Thanks!

If you compare using:
if ('0' == 0) // if '0' is equal to 0
it should return true as the values are compared with the string being converted to a number. If you do:
if ('0' === 0) // if '0' is identical to 0
it will return false as they have to be of the same type too.
Note the triple '='

You can also force their type to be the same before comparing:
if((int)'0' === (int)0) {
// true
}
if((string)'0' === (string)0) {
// true
}

Related

Very strange behaviour in PHP [duplicate]

This question already has answers here:
Comparing String to Integer gives strange results
(5 answers)
Closed 6 years ago.
I have this code:
$test = 0;
if ($test == "on"){
echo "TRUE";
}
Result of this code will be:
TRUE
WHY??? My version of PHP: 5.4.10.
If you compare a number with a string or the comparison involves numerical strings, then each string is converted to a number and the comparison performed numerically. These rules also apply to the switch statement. The type conversion does not take place when the comparison is === or !== as this involves comparing the type as well as the value.
$test = 0;
if ($test === "on"){
echo "TRUE";
}
PHP will convert the string to number to compare. Using ===, will compare the value as well as the type of data.
var_dump(0 == "a"); // 0 == 0 -> true
var_dump("1" == "01"); // 1 == 1 -> true
var_dump("10" == "1e1"); // 10 == 10 -> true
var_dump(100 == "1e2"); // 100 == 100 -> true
Docs
Because you are comparing $test with string value not binary value, if you want to compare with string value, try with === comparison, for
value + dataType.
In your example, var_dump(0=="on"); always return bool(true).
But when you use var_dump(0==="on"); it will give you bool(false).
Example from PHP Manual:
var_dump(0 == "a"); // 0 == 0 -> true
var_dump("1" == "01"); // 1 == 1 -> true
var_dump("10" == "1e1"); // 10 == 10 -> true
var_dump(100 == "1e2"); // 100 == 100 -> true
If you compare a number with a string or the comparison involves numerical strings, then each string is converted to a number and the comparison performed numerically. These rules also apply to the switch statement. The type conversion does not take place when the comparison is === or !== as this involves comparing the type as well as the value.
So use "===" for comparision.
Refer the link : http://php.net/manual/en/language.operators.comparison.php
This is because your are doing ==
0 is integer so in this on converted to int which is 0
So your if statement looks like 0==0
For solution You have to use ===
if ($test === "on")
In PHP, loose comparison between integer and non-numeric string (i.e string which cannot be interpreted as numeric, such as "php" or in your case "on") will cause string to be interpreted as integer with value of 0. Note that numeric "43", "1e3" or "123" is a numeric string and will be correctly interpreted as numeric value.
I know this is weird.

php variables comparison == vs ===

this is code:
$s = 0;
$d = "dd";
if ($s == $d) {
var_dump($s);
die(var_dump($d));
}
result is:
int 0
string 'dd' (length=2)
Please explain why.
why ($s == $d) results as true?
Of course, if === is used it will results as false but why this situation requires ===?
Shouldn't it be returned false in both situations?
Because (int)$d equals with 0 and 0=0
you must use strict comparison === for different character tyes (string) with (int)
Your $d is automatically converted to (int) to have something to compare.
When you compare a number to a string, the string is first type juggled into a number. In this case, dd ends up being juggled into 0 which means that it equates to true (0==0).
When you change the code to:
<?php
$s = 1;
$d = "dd";
if ($s == $d)
{
var_dump($s);
die(var_dump($d));
}
?>
You will find that it doesn't pass the if statement at all.
You can more details by reading up on comparison operators and type juggling.
The string "dd" is converted to int, and thus 0.
Another example :
if ( "3kids" == 3 )
{
return true;
}
And yes, this returns true because "3kids" is converted to 3.
=== does NOT auto convert the items to the same type.
Also : 0 == false is correct, but 0 === false is not.
See : http://php.net/manual/en/language.types.type-juggling.php
The string will try to parsed into a number, returns 0 if it is not in right number format.
As seen in the php website :
http://php.net/manual/en/language.operators.comparison.php
var_dump(0 == "a"); // 0 == 0 -> true
In PHP, == should be pronounce "Probably Equals".
When comparing with ==, PHP will juggle the file-types to try and find a match.
A string with no numbers in it, is evaluated to 0 when evaluated as an int.
Therefore they're equals.

PHP: If statement issue

The array resAlloc contains 10 columns and 5 rows. All entries are equal to 0. So, I expect the following IF statement be TRUE, but it's false for some reason... Why?
if ($resAlloc[$i][$j] != 'ts' && $resAlloc[$i][$j] != 't' && $resAlloc[$i][$j] != 'st') {
$count++;
}
!= evaluates 0 as false. Use !== which is more strict.
IIRC anything equaling 0 with != will return FALSE.
The problem is that you're comparing a string and an integer, and PHP is "helpfully" casting the string to an integer -- the integer zero. 0!='ts' evaluates as false, because the comparison it ends up doing after conversion is 0!=0. You can prevent this by explicitly treating the contents of your array as a string:
strval($resAlloc[$i][$j]) != 'ts'
This will do the comparison '0'!='ts', which correctly evaluates to true. If you pass strval() a string it returns it unchanged, so this should be safe to use regardless of what's in your array.
Alternately, as Samy Dindane said, you can just use !== which won't do any type conversion.

PHP Type-Cast Confusion

I have the following code:
<?php
$val = 0;
$res = $val == 'true';
var_dump($res);
?>
I always was under impression that $res should be 'false' as in the above expression PHP would try to type cast $val to boolean type (where zero will be converted as false) and a string (non-empty string is true). But if I execute the code above output will be:
boolean true
Am I missing something? Thanks.
In PHP, all non-empty, non-numeric strings evaluate to zero, so 0 == 'true' is TRUE, but 0 === 'true' is FALSE. The string true has not been cast to a boolean value, but is being compared as a string to the zero. The zero is left as an int value, rather than cast as a boolean. So ultimately you get:
// string 'true' casts to int 0
0 == 0 // true
Try this:
echo intval('true');
// 0
echo intval('some arbitrary non-numeric string');
// 0
Review the PHP type comparisons table. In general, when doing boolean comparisons in PHP and types are not the same (int to string in this case), it is valuable to use strict comparisons.
Because $val is the first operator PHP converts the string true to an integer which becomes 0. As a result 0 ==0 and your result is true;
Try this
<?php
$val = 1;
$res = (bool)$val == 'true';
var_dump($res);
?>

Not equal to != and !== in PHP

I've always done this: if ($foo !== $bar)
But I realized that if ($foo != $bar) is correct too.
Double = still works and has always worked for me, but whenever I search PHP operators I find no information on double =, so I assume I've always have done this wrong, but it works anyway. Should I change all my !== to != just for the sake of it?
== and != do not take into account the data type of the variables you compare. So these would all return true:
'0' == 0
false == 0
NULL == false
=== and !== do take into account the data type. That means comparing a string to a boolean will never be true because they're of different types for example. These will all return false:
'0' === 0
false === 0
NULL === false
You should compare data types for functions that return values that could possibly be of ambiguous truthy/falsy value. A well-known example is strpos():
// This returns 0 because F exists as the first character, but as my above example,
// 0 could mean false, so using == or != would return an incorrect result
var_dump(strpos('Foo', 'F') != false); // bool(false)
var_dump(strpos('Foo', 'F') !== false); // bool(true), it exists so false isn't returned
!== should match the value and data type
!= just match the value ignoring the data type
$num = '1';
$num2 = 1;
$num == $num2; // returns true
$num === $num2; // returns false because $num is a string and $num2 is an integer
$a !== $b TRUE if $a is not equal to $b, or they are not of the same type
Please Refer to http://php.net/manual/en/language.operators.comparison.php
You can find the info here: http://www.php.net/manual/en/language.operators.comparison.php
It's scarce because it wasn't added until PHP4. What you have is fine though, if you know there may be a type difference then it's a much better comparison, since it's testing value and type in the comparison, not just value.

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