PHP in_array not working - php

I am using the PHP in_array() function in order to authenticate (with sessions) if a user can access a particular page. For some reason, it is not working...
PHP PAGE
session_start();
require_once('../scripts/functions.php');
$role_auth = #$_SESSION['role_auth'];
access($role_auth, array(0,1,2,3,4));
access FUNCTION
function access($role_auth, $array){
if(!(in_array($role_auth, $array))){
header("Location: ../index.html");
}
}
If I insert print statements in the function, I can see that all of the correct values are being passed into the function. The problem is, if the function is called without a session variable set, for some reason it is considered as being in the array, and it authenticates.
Any ideas?

you may want to enable strict type checks by using:
in_array($role_auth, $array, true)
as what is likely happening is that $role_auth is being eval'd as false and that could match 0 in your in_array statement.
what you SHOULD be doing is this:
session_start();
require_once('../scripts/functions.php');
$role_auth = (isset($_SESSION['role_auth']))?$_SESSION['role_auth']:-1;
access($role_auth, array(0,1,2,3,4));
or something similiar. nothing good ever comes of using the # operator

I would check to see if $_SESSION['role_auth'] is actually set (with isset) instead of using # to suppress warnings (which is bad practice IMHO)
I think what's happening is that false == 0 ... so in_array returns true when nothing is in $role_auth because it sees it as 0 and 0 is in your array

$role_auth = #$_SESSION['role_auth'];
The # sign is suppressing any warnings you might get here, like index is not in array. How about something like this instead:
if(isset($_SESSION['role_auth']))
$role_auth = $_SESSION['role_auth'];
else
$role_auth = -1;//(or whatever an invalid role is)

In php, the number zero is considered equal to most non-numeric things, for example:
null == 0
false == 0
"" == 0
"asdf" == 0
You probably need to make sure that $_SESSION actually contains the 'role_auth' key beforehand and convert it to the appropriate type, also passing the $strict parameter to in_array, thus guaranteeing a type check as well as a value check (=== vs. ==). Removing zero from your array might also be a good idea.

Related

Check if variable has value and isn't empty

I want to check if a numeric variable has a value (including '0') and is not empty. Empty meaning EMPTY (''), not '0'.
Is this really the best I can do with PHP?
if (isset($variable) && $variable !== '') { ... }
I'd like to do this with one check without writing a function for it...
What you are trying to check is string length, not "empty". This can easily be done using strlen().
if (isset($variable) && strlen($variable) > 0) {
// Do something
}
If you want to exclude whitespace as invalid, you can add a trim() in there as well (generally recommended).
if (isset($variable) && strlen(trim($variable)) > 0 } {
// ...
}
The best thing you could do, is making your own custom function. The point is to pass the variables by reference to not trigger a warning, when you pass an undefined variable. As posted as comment, I'd use something along the line isset($variable) AND !empty($variable) AND !is_numeric($variable) AND $variable !== false to cover all cases.
Your custom function could look like this (improved version):
function is_blank(&$variable) {
return (bool) !(isset($variable) AND (!empty($variable) OR is_numeric($variable) OR $variable === false));
}
https://3v4l.org/ZcCDu
Yes, your way is the best (most efficient) way to:
insure the variable has been set (so you don't get an warning checking a variable that's not been set)
it's not the empty string ''
But, could be '0', 0,false, null, or [] which all count as empty in php, but you wish to consider as non-empty as indicated by your OP
your !== will ensure only exactly the string '' is compared (no casting/conversion)
The use of strlen works as well, but if you look at the opcode generated you'll see direct comparison is more 3 times computationally more efficient (assuming all operations are equally weighted, even more efficient if operations like DO_FCALL take significantly more cycles to execute than a basic IS_NOT_IDENTICAL check)
The !== ''version bytecode:
IS_NOT_IDENTICAL ~1 !0, ''
The strlen() > 0 version bytecode:
SEND_VAR !0
DO_FCALL 1 $1 'strlen'
IS_SMALLER ~2 $1, 0
(The answer has been edited. Consult the additionals further down under "ternary operations").
Why go through the trouble of using all that?
Just use an "not empty" if(!empty($var)){...}
However, if you're using this with a GET array, then yes; it would be best to use an isset() and empty() on a conditional statement.
I want to check if a variable has a value (including '0') and is not empty
That to me interprets as:
Check if a value has a value and is not empty (as you wrote) and stands to contain a 0 (zero).
Therefore:
if(!empty($var) && $var='0'){...}
I'd like to do this with one check without writing a function for it...
Use a ternary operator then.
However "without a function"... right well you can't. You still need "some type of function".
About that "ternary operator" I mentioned above. You can reference what are called "nested ternary operations" in both these Q&A's on Stack:
How to concatenate multiple ternary operator in PHP?
nested php ternary trouble: ternary output != if - else
That way you won't need a custom function.
Sidenote: I am by far not taking away or trying to take away from (Charlotte's) accepted answer (which should remain as accepted). This is just an additional method of achieving your (ultimate) goal.

Simplified test for non-empty variable

I am maintaining some old PHP code and I find many places in the code a test for a variable being non-empty of the following form:
if (!(isset($field_name) && strlen($field_name) > 0))
To my way of thinking, the following much simpler form would do exactly the same thing:
if ($field_name)
Am I missing some subltety whereby the more complex form is more precise? I have tended to use the simpler form any place where I need to add new code.
You can use empty() to replace your first line:
if (!empty($field_name))
The problem with your second example is that it will generate a warning if the variable is not set. Both empty() and isset() will not generate a warning for non-existing variables.
Note that you always have to account for possible values so if your value can be 0 or '0', this will not work as after $var = 0;, empty($var) will evaluate to true.
if ($field) will fail if $field is null, 0, 0.00, false or anything that can be casted false
if(!isset($field))
will fail if $field has not been declared at all
They do not do the same thing. The first code sample:
Checks if $field_name exists (I figure the ! is unintentional as it doesn't make sense the way it is written)
Checks if $field_name has a string length greater than zero
The second code sample:
Checks if the variable has a Boolean true value
Will throw an error if $field_name is not set.
The first snippet is clear and precise in its intent and performs a specific task.
The second snippet is very basic and only verifies the variable has a Boolean true value (and there are better ways to do that). If $field_name contains a string "0" this snippet will cause a hard to spot bug as it will fail as "0" evaluates to false. The first snippet would catch this.
You are right that the code you found is odd. It should probably be either:
if(!(isset($field_name)) { ... do something }
or
if(isset($field_name) && strlen($field_name) > 0 ) { ... do something }
...As you can appreciate, there is no need to test the length of a variable that is not defined.
However, if($field_name) is not the same as if(!(isset($field_name)), and the difference is not subtle. Indeed, the former will earn you a Undefined variable: fieldname if by some stroke of bad luck $field_name is not defined.
What's the difference?
if($field_name) tests if the existing variable $field_name evaluates to TRUE. For instance, it's value might be "my dog", and that evaluates to TRUE
if(!(isset($field_name)) tests if the variable $field_name exists at all.

PHP if (variable)

In various PHP tutorials I see this syntax -
if ($_POST) {
do something
}
I want to know whether this is equivalent to either isset or !(empty) (either one) or has different properties.
It attempts to evaluate the expression and cast it to boolean.
See 'Converting to boolean' at http://php.net/manual/en/language.types.boolean.php to see which values will be equivalent to true and which to false.
It does NOT however check for array key existence (i.e. isset), so if you try if ($_GET['somekey']) but somekey does not exist, you will see PHP Notice: Undefined index: somekey and then false will be assumed.
The best practice would be to perform empty() or isset() checks manually first as fits.
Good question. You are adressing one of PHPs dark sides if you ask me.
The if statement
Like in any other language I can imagine if evaluates the parameter to either true or false.
Since PHP doesn't really know types you could put any expression as parameter which will then be casted to bool as a whole
Following values are considered to be "FALSE"
boolean FALSE
integer 0
float 0.0
empty string
string "0"
any array with zero elements
NULL e.g. unset variables or $var = null
SimpleXML objects when created from empty tags
EVERY other value or expression result is casted to bool TRUE
Now, knowing this, all we need to find out is, what an expression or function returns when executed
If no POST data is set, the following expression would be TRUE
$_POST == FALSE
The isset function
isset returns bool TRUE when the given variable is set and not null.
parameters can be variables, array elements, string offsets and data members of objects.
In PHP 5.4 they fixed the behaviour with string offsets
$var = FALSE;
isset( $var ) === TRUE;
$var === FALSE;
More here
http://de1.php.net/manual/en/function.isset.php
The empty function
Returns false when a variable is considered to be empty or does not exist.
Those values are considered empty:
Returns FALSE if var exists and has a non-empty, non-zero value. Otherwise returns TRUE.
The following values are considered to be empty:
"" (empty string)
0 (integer)
0.0 (float)
"0" (string)
NULL
FALSE
array() (empty array)
Also declared variables without value are empty
compare table
$var = FALSE;
isset($var) === TRUE;
empty($var) === TRUE;
$var === FALSE;
if ($_POST)
This will evaluate to true if there are any elements in the POST array.
if(isset($_POST))
This will always evaluate to true because the POST array is always set, but may or may not contain elements, therefore it is not equivalent to the first example.
if(!empty($_POST))
This however, is equivalent to the first example because empty() checks for contents in the array.
A good generic way of testing if the page was posted to is:
if($_SERVER['REQUEST_METHOD'] == 'POST')
$_POST:
this is used to find whether data is passed on using HTTP POST method and also extracting the variables sent through the same which are collected in an associative array
isset:
checks whether a variable is set(defined) or is NULL(undefined)
PHP.net is an invaluable source of information for figuring out the intricacies and quirks of the language.
With that said, those are not equivalent. The if statement converts the expression to a boolean value (see here for information on what is considered false).
isset is used to "determine if a variable is set and is not NULL."
empty determines whether a variable is empty, i.e., "it does not exist or if its value equals FALSE."
Best practice if you want to check the value of a variable but you aren't sure whether it is set is to do the the following:
if(isset($var) && $var) { ... }
ie check isset() AND then check the variable value itself.
The reason for this is that if you just check the variable itself, as per the example in your question, PHP will throw a warning if the variable is not set. A warning message is not a fatal error, and the message text can be suppressed, but it's generally best practice to write code in such a way that it doesn't throw any warnings.
Calling isset() will only tell you whether a variable is set; it won't tell you anything about the value of the variable, so you can't rely on it alone.
Hope that helps.

PHP if() evaluation problem needs a rewrite

I noticed this weird evaluation yesterday after searching for a few hours in my code for an error. i am passing scores into php, sometimes the score=0 which causes an issue.
send php ?blah=blah&score=0
if(!empty($_REQUEST['score']){
//do database update stuff
}else{
// show entire webpage
}
It works great unless the score=0 the if() will evaluate to false and return the entire webpage to my ajax handler and error. I have temporarily changed !empty to isset but this will cause problems in the future because isset evaluates to true even if the score key is in the url string without a value.
ex: (?blah=blah&score=&something=else)
my question is: what is the best way to recode this to work correctly now and in the future?
edit: there are a few working answers here, i appreciate everyones time. it was difficult to choose an answer
As the manual says, a variable is considered empty() if it has an empty or zero value.
So it will treat your variable wrongly as empty even though 0 is a perfectly acceptable value in your case.
If you need score to be a number, you could use isset() in combination with a is_numeric() check instead:
if((isset($_REQUEST['score']) and (is_numeric($_REQUEST['score'])){
Check out the manual page to see the kinds of values is_numeric() accepts. If score is always an integer, you can also use is_int((int)$_REQUEST['score']) but that will convert invalid input values to 0.
Additionally, as #sightofnick says, it's better to use explicit $_GET or $_POST instead of $_REQUEST.
Re your update:
In that case I would
Do check whether the variable is "0" (string "zero")
If it is "0", make it 0 (integer "zero")
If it is not 0, convert it to an integer (int)$_REQUEST["score"])
If the conversion resulted in 0, it was invalid input - exit
You have a valid integer variable.
empty() will return false if a value is zero. Use isset() or array_key_exists() instead, if you want to check if a variable in an array is set:
if (array_key_exists('score', $_REQUEST)) {...}
Try doing
if (isset($_REQUEST['score']) && ($_REQUEST['score'] !== '')) {
...
}
The isset will handle the presence/absence of the query parameter, and the strict string (!==) comparison will handle the case where the 'score' query is present but has no value. PHP treats all data coming from _GET/_POST/_REQUEST as strings, so this test is 100% reliable.
if(isset($_REQUEST['score']) && $_REQUEST['score'] != ''){
//do database update stuff
}else{
// show entire webpage
}
You may be able to solve that with
if (isset($_REQUEST['score']) && is_numeric($_REQUEST['score'])) {}
That of course if scrore can only contain numeric value

Is this an OK test to see if a variable is set

Yesterday, I posted an answer to a question that included several (unknown to me at the time) very bad code examples. Since then, I've been looking at my fundamental knowledge of PHP that allowed me to think that such code is possible. This brings me to a question that I can't seem to find an answer to:
If I want to check for whether or not a variable has anything set, is it a valid practice to not use isset() or another helper function? here's a "for instance":
if($not_set){
//do something
} else {
//do something else
}
Rather than...
if(isset($not_set)){
//do something
} else {
//do something else
}
From the name of the variable, you can see that this variable is not set. Therefore the conditional would be false and the else portion would run. Up until now I have been using this practice, but after the posts yesterday, I now have an inkling that this is wrong.
Here's why I thought that it would be an ok practice to leave out the isset() function above. From PHP manual:
The if construct is one of the most
important features of many languages,
PHP included. It allows for
conditional execution of code
fragments. PHP features an if
structure that is similar to that of
C:
if (expr) statement
As described in the section about
expressions, expression is evaluated
to its Boolean value. If expression
evaluates to TRUE, PHP will execute
statement, and if it evaluates to
FALSE - it'll ignore it. More
information about what values evaluate
to FALSE can be found in the
'Converting to boolean' section.
And from the 'Converting to boolean section':
When converting to boolean
, the following values are considered
FALSE:
...
* the special type NULL (including unset variables)
Why would the manual go out of its way to state that unset variables are included if this is a bad practice? If it's unset, it gets converted to NULL and therefore is evaluated properly by the conditional. Using isset() will find the same result, but will take extra cycles to do so.
Have I been wrong this whole time, and if so, why? (And just how bad it is, maybe?)
If the variable is not set you get a Notice. If you use isset() you don't get a notice. So from an error reporting point of view, using isset() is better :)
Example:
error_reporting(E_ALL);
if($a) {
echo 'foo';
}
gives
Notice: Undefined variable: a in /Users/kling/test on line 5
whereas
error_reporting(E_ALL);
if(isset($a)) {
echo 'foo';
}
does not output anything.
The bottom line: If code quality is important to you, use isset().
It's okay but not good practice to use if to check for a set variable. Two reasons off the top of my head:
Using isset makes the intent clear - that you're checking whether the variable is set, and not instead checking whether a condition is true.
if ($not_set) will evaluate to false when $not_set is actually set but is equal to boolean false.
You will run in to problems if your variable is set, but evaluates to FALSE, like the following:
the boolean FALSE itself
the integer 0 (zero)
the float 0.0 (zero)
the empty string, and the
string "0"
an array with zero elements
an object with zero member
variables (PHP 4 only)
the special type NULL (including
unset variables)
SimpleXML objects created from empty
tags
Taken from the PHP manual.
Basically, using isset() is showing that you are explicitly checking if a variable exists and is not NULL, while the structure of your if statement only checks if the variable is true. It is more clear and less error-prone.
It is a common practise, but is not good -- you should always use isset!
If your $not_set is set, and is a bool with the value false, your "test" will fail!
isset works as a guard preventing you from using variables that do not actually exist.
if (isset($foo)) and if ($foo) do not mean the same thing. isset just tells you if the variable actually exists and if it's okay to use it, it does not evaluate the value of the variable itself*.
Hence, you should usually use one of these two patterns:
If the variable is sure to exist and you just want to check its value:
if ($foo == 'bar')
If the variable may or may not exist, and you want to check its value:
if (isset($foo) && $foo == 'bar')
If you're just interested that a variable is set and evaluates to true, i.e. if ($foo), you can use empty:
if (isset($foo) && $foo)
// is the same as
if (!empty($foo))
* it does check for null, where null is as good as not being set at all

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