I have a database with some football scores (goal differences).
p12_score1 contains 3, -2, 0 and NULL
All I want to do is to print all score differences (including 0) and ignore the row that has no value (NULL in mysql). However is_null also considers the value 0 a null and won't print it.
IF(!$p12_score1==NULL){
$equal[] = $p12_score1;
}
else {
echo "null";
}
print_r ($equal);
is_null does not consider the value 0 equal to null.
Your code uses == for the comparison, which does not make this distinction. Use the identical operator === or is_null instead:
// One way to do it:
if(!$p12_score1 !== NULL)
// Another:
if(!is_null($p12_score1))
However is_null also considers the value 0 a null and won't print it.
No. http://codepad.org/mjiGfhTB
However, you use the comparisong operator
$p12_score1==NULL
and in this case: In loosely typed languaged null is equal to 0 (and some others ;)). However, PHP supports the identity-comparisong
$p12_score1===NULL
A sidenote: !$a == $b is slightly ugly. Use the real "not equal" !=, or !==.
Try this:
IF(!empty($p12_score1)){
$equal[] = $p12_score1;
}
else {
echo "null";
}
print_r ($equal);
Try
if($p12_score1 !== null) {
$equal[] = $p12_score1;
} else {
echo "null";
}
print_r ($equal);
If you want to do this in PHP instead of sql you have to change your code to this:
IF (!$p12_score1 === NULL)
This way you check that $p12_score1 is exactly null.
The function is_null should work. Another option is is_int. I tried both and both work correctly.
Related
if ($user_id == NULL || $user_name == NULL || $user_logged == NULL) {
$user_id = '-1';
$user_name = NULL;
$user_logged = NULL;
}
if ($user_admin == NULL) {
$user_admin = NULL;
}
Is there any shortest way to do it ?
And if i right, it should be tested with is_null?
It's possible $user_id, $user_name and $user_logged write in one line (maybe array?) without repeating NULL ?
If you want to test whether a variable is really NULL, use the identity operator:
$user_id === NULL // FALSE == NULL is true, FALSE === NULL is false
is_null($user_id)
If you want to check whether a variable is not set:
!isset($user_id)
Or if the variable is not empty, an empty string, zero, ..:
empty($user_id)
If you want to test whether a variable is not an empty string, ! will also be sufficient:
!$user_id
You can check if it's not set (or empty) in a number of ways.
if (!$var){ }
Or:
if ($var === null){ } // This checks if the variable, by type, IS null.
Or:
if (empty($var)){ }
You can check if it's declared with:
if (!isset($var)){ }
Take note that PHP interprets 0 (integer) and "" (empty string) and false as "empty" - and dispite being different types, these specific values are by PHP considered the same. It doesn't matter if $var is never set/declared or if it's declared as $var = 0 or $var = "". So often you compare by using the === operator which compares with respect to data type. If $var is 0 (integer), $var == "" or $var == false will validate, but $var === "" or $var === false will not.
here i have explained how the empty function and isset works please use the one that is appropriate also you can use is_null function also
<?php
$val = 0;
//evaluates to true because $var is empty
if (empty($val)) {
echo '$val is either 0, empty, or not set at all';
}
//evaluates to true because $VAR IS SET
if (isset($val)) {
echo '$val is set even though it is empty';
}
?>
empty() is a little shorter, as an alternative to checking !$user_id as suggested elsewhere:
if (empty($user_id) || empty($user_name) || empty($user_logged)) {
}
To check for null values you can use is_null() as is demonstrated below.
if (is_null($value)) {
$value = "MY TEXT"; //define to suit
}
Please define what you mean by "empty".
The test I normally use is isset().
you can use isset() routine .
also additionaly you can refer an range of is_type () functions like
is_string(), is_float(),is_int() etc to further specificaly test
1.
if(!($user_id || $user_name || $user_logged)){
//do your stuff
}
2 . No. I actually did not understand why you write such a construct.
3 . Put all values into array, for example $ar["user_id"], etc.
<?php
$nothing = NULL;
$something = '';
$array = array(1,2,3);
// Create a function that checks if a variable is set or empty, and display "$variable_name is SET|EMPTY"
function check($var) {
if (isset($var)) {
echo 'Variable is SET'. PHP_EOL;
} elseif (empty($var)) {
echo 'Variable is empty' . PHP_EOL;
}
}
check($nothing);
check($something);
check($array);
Its worth noting - and I only found this out after nearly 9 years of PHP coding that the best way of checking any variable exists is using the empty() function. This is because it doesn't generate errors and even though you turn them off - PHP still generates them! empty() however won't return errors if the variable doesn't exist. So I believe the correct answer is not to check if its null but to do the following
if (!empty($var) && is_null($var))
Note the PHP manual
variable is considered empty if it does not exist or if its value equals FALSE
As opposed to being null which is handy here!
Felt compelled to answer this because of the other responses. Use empty() if you can because it covers more bases. Future you will thank me.
For example you will have to do things like isset() && strlen() where instead you could use empty(). Think of it like this empty is like !isset($var) || $var==false
The best and easiest way to check if a variable is empty in PHP is just to use the empty() function.
if empty($variable)
then
....
I've been working on a very basic search engine. It basically operates by checking if the word exists. If it does, it returns the link. I know most of you would suggest to create a database from phpMyAdmin but I don't remember the password to make the mySql_Connect command work.
Anyway here is the code:
<?php
session_start();
$searchInput = $_POST['search'];
var_dump($inputPage1);
var_dump($searchİnput);
$inputPage1 = $_SESSION['pOneText'];
$inputPage2 = isset($_SESSION['pTwoText']) ? $_SESSION['pTwoText'] : "";
$inputPage3 = isset($_SESSION['pThreeText']) ? $_SESSION['pThreeText'] : "";
if (strpos($inputPage1, $searchInput)) {
echo "True";
} else {
echo "False";
}
?>
When I search a word, any word from any page, weather it exists or not, it always returns false. Does anyone know why?
From the PHP documentation:
Warning: This function may return Boolean FALSE, but may also return a non-Boolean value which evaluates to FALSE. Please read the section on Booleans for more information. Use the === operator for testing the return value of this function.
So the function returns the integer 0 since $searchInput starts at the first character of $inputPage1. Since it is inside an if condition, that expects a boolean, the integer is then converted to one. When converted to boolean, zero is equal to false so instead the else block is executed.
To fix it, you need to use the !== operator (the not equal equivalent of ===):
if (strpos($inputPage1, $searchInput) !== false) {
//...
Try stripos() to match case insensitive
First print all items in $_POST and $_SESSION using
echo "<pre>";
print_r($_POST);
print_r($_SESSION);
and ensure that the search string really exist in the bigger string .
Also make sure that your are using "false" to compare :
i.e
$pos = strpos($biggerString,$seachString);
if($pos !== false)
{
echo "Not found";
}
Is there a diference between this comparisons ?
What is the diference between ! and === FALSE ?
if (!class_exists($class)) {
require($class.'.php');
}
if (class_exists($class) === FALSE) {
require($class.'.php');
}
In this case, no.
Some people think it's good programming style to explicitly show that they're comparing to a boolean. Personally... I don't like it, but I guess the more verbose form is more obvious, as the ! operator isnt the mose visible thing when smashed between a parenthesis and other vertically'ish characters.
Yes both are the different things:
php automatically considers 0 as "false" and 1 as "true" so when ever you use function response directly inside the if condition at that this both makes a difference.
consider a function, if executed properly at that it returns int number. it may be 0 too.
But if function did not match requirement at that it is returning false.
So at this time function returning value 0 is success. event though the result is zero. At this if you check this in if condition like
$return = someFunction();
if($return){
//code if ture
}
so if $return is 0 your if code will not be executed even your function execution was correct so in that case you should check like
$return = someFunction();
if($return !== FALSE){
//code if ture
}
=== and !== are used to check the response exactly match return type also.
if('0' === 0)
will return false
but
if('0' == 0)
will return true...
Hope your idea is clear now.
check this out:
if('0' == 0){
echo 'Hi, I will be in screen :)';
}
if('0' === 0){
echo 'I will not be in screen :(';
}
Currently whenever i am trying to determine when a variable is not defined i use the code
if($variable=="")
however i have been told by people to use the function
if(empty($variable))
Reading on this, it returns false if the value is 0, and i have plenty of array values that are zero that cant be returning false. i could always add ||$variable==0) to skip this.
But all i am asking is why is this a preferred method for determining empty variables, is it efficiency or is there more to it than that?
If you need to know, use strict type comparison:
if (empty($variable) && $variable !== 0) // !== instead of !=
Generally, you should use isset() - if the variable hasn't yet been defined, it returns false:
$a = 4;
echo isset($a) ? 'a' : 'no a'; // a
echo isset($b) ? 'b' : 'not to b'; // not to b
I want to check if a query variable exists or not. Then I would do something based on that query value. If it exists and is true, do something. If it doesn't exist or is false, do something else such as show a 404 page.
e.g If the url was domain.com?konami=true
if (condition) {
//something
} else {
//show404
}
OPs question is a bit unclear. If you assume that he wants to check that konami is a $_GET parameter and that it has the value of "true" do:
if (isset($_GET["konami"]) === true && $_GET["konami"] === "true") {
// something
} else {
// show 404
}
The problem with the current accepted answer (by Cameron) is that it's lacking the isset check (which is unforgivable, it is objectively wrong). The problem of the highest voted answer (by Jan Hancic) is that it lacks the === "true" check (which is debatable, it depends on how your interpret the question).
Note that && is a lazy-and, meaning that if the first part is false, the second part will never be evaluated, which prevents the "Undefined index" warning. So the order of the statements is important.
Also note that $a === true is not the same as $a === "true". The first compares a boolean whereas the second compares a string.
If you do weak comparison $a == true you are checking for truthy-ness.
Many values are truthy, like the string "true", the number 1, and the string "foo".
Examples of falsy values are: empty string "", the number 0 and null.
"true" == true; // true
"foo" == true; // true
1 == true; // true
"" == true; // false
0 == true; // false
null == true; // false
"true" === true; // false
"true" === false; // false
There is a little confusion around what value should be tested. Do you want to test the konami parameter for being true in the sense of boolean, i.e. you want to test konami parameter for being truthy, or do you want to test if it has string value equal to "true"? Or do you want to test konami parameter for any value in general?
I guess what is wanted here is to test konami for a given string value, "true" in this case, and for being set at the same time. In this case, this is perfectly enough:
ini_set('error_reporting', E_ALL & ~E_NOTICE);
...
if ($_GET['konami'] == "true")
...
This is enough, because if the $_GET['konami'] is unset, it cannot be equal to any string value except for "". Using === is not neccessary since you know that $_GET['konami'] is string.
Note that I turn off the E_NOTICE which someone may not like - but these type of "notices" are normally fine in many programming languages and you won't miss anything if you disable them. If you don't, you have to make your code unecessarily complex like this:
if (isset($_GET['konami']) && $_GET['konami'] == "true")
Do you really want to complicate your code with this, or rather make it simple and ignore the notices like Undefinex index? It's up to you.
Problems with other answers as you mentioned:
#Jan Hancic answer: it tests for true, not "true".
#Cameron answer: might be simplified and he didn't mention the necessity of disabling E_NOTICE.
#Frits van Campen's answer: too complex to my taste, unnecessary test for === true
Umm this?
if (isset($_GET['konami']) === true) {
// something
} else {
//show 404
}
Easy:
if(isset($_GET['konami']) && $_GET['konami'] != 'false') {
//something
} else {
// 404
}
quick and simple.
$konami = filter_input(INPUT_GET, 'konami', FILTER_VALIDATE_BOOLEAN) or die();
ref:
filter flags
filter_input
You may try this code. In this code checked two conditions by one if condition that is $konami contains value and $konami contains 'true'.
$konami = $_GET['konami'];
if( ($konami) && ($konami == "true")){
/*enter you true statement code */
}else {
/* enter your false statement code */
}
You can do it like this:
$konami = false;
if(isset($_GET['konami'])){
$konami = $_GET['konami'];
}
if($konami == "true"){
echo 'Hello World!';
}
else{
header('HTTP/1.0 404 Not Found');
}
In this case you'll always have $konami defined and - if set - filled with the value of your GET-parameter.
if(!$variable) {
//the variable is null
die("error, $variable is null");
}else{
//the variable is set, your code here.
$db->query("....");
}
This works best:
$konami = $_GET['konami'];
if($konami == "true")
{
echo 'Hello World!';
}
else
{
header('HTTP/1.0 404 Not Found');
}
Easiest and shortest way of doing it:
if($konami != null){ echo $konami; } else { header('HTTP/1.0 404 Not Found'); }