I have a problem with my php script it is...
Parse error: syntax error, unexpected T_LOGICAL_OR, expecting ')' in C:\wamp\www\register.php on line 34
I am a nube with !Empty so let me know if im doing something wrong.
if (!empty($username or $email or $password or $repassword))
{
}
else
{
}
This is my code but something in the middle of the else and if statement
So what am i doing wrong?
My question is what am i doing wrong?
oh and
$username
$email
$password
$repassword
are all values
and yes this is a script to figure out if someone has filled in all of the textboxes in the form for a register script.
Documentation Section says
empty() only checks variables as anything else will result in a parse
error. In other words, the following will not work:
empty(trim($name)).
empty is only able to check a variable for its emptiness. $username or $email or … is a boolean expression which either evaluates to TRUE or FALSE – not what you want. you have to use multiple empty calls:
if (!(empty($username) or empty($email) or empty($password) or empty($repassword))) {
// all set!
}
It sounds like you want something along these lines...
if (!empty($username) && !empty($email) && !empty($password) && !empty($repassword))
{
// none of the values are empty
}
else
{
// at least one of the values is empty
}
empty() will check whether or not a variable which is passed as its argument is considered to be empty, which means it will check whether or not its argument contains -
"" (an empty string)
0 (0 as an integer)
0.0 (0 as a float)
"0" (0 as a string)
NULL
FALSE
array() (an empty array)
var $var; (a variable declared, but without a value in a class)
In your code, you are providing $username or $email or $password or $repassword as its argument. This is a boolean expression, meaning it will evaluate to either TRUE or FALSE. So the argument to empty() will be a boolean value, not a variable. This is the reason why you are getting this error.
I am assuming that you want to check whether or not all of these variables are empty. If this is the case, then you can do this in the following way -
if( !empty($username) or
!empty($email) or
!empty($password) or
!empty($repassword))
{
..........
}
else
{
........
}
You can't use the OR operator for calling a function multiple times with multiple variables.
If you want to make sure that all variables are not empty, use this:
if ( ! (empty($username) ||
empty($email) ||
empty($password) ||
empty($repassword)
)
)
Empty only works on variables
http://php.net/manual/en/function.empty.php
The expression x or y or z will be evaluated and therefore will not be a variable.
This won't work because empty() can only check a single variable.
Even if it did work, you are doing it wrong, this if statement would check if any of the four variables is not empty. Therefore - if it was valid code - it would return true even if only one variable was defined.
So change it to this:
if (!empty($username) and !empty($email) and !empty($password) and !empty($repassword))
You can't have multiple OR's in one empty function. Your code will have to be like this:
if (!empty($strValue1) OR !empty($strValue2) OR !empty($strValue3)) {
// code
} else {
// other code
}
Related
I found an example php assignment statement online which maybe resembles a tertary conditional condensed statement, but not quite. Does anyone have insight as to how to read this assignment statement?
$access = $access || $note->uid == $user->uid && user_access('note resource view own notes');
My first guess was "assign to access whatever is in access, or if empty, whether uid values are equal and'd with the return of user_access." But I get the feeling that is not correct, as it seems illogical.
First have a look at the Operator Precedence
== comes before && comes before || comes before =
Thus your statement is more clear with adding the following parentheses:
$access = (
$access
||
(
($note->uid == $user->uid)
&&
user_access('note')
)
);
assign to access whatever is in access, or if empty,
Not quite: assign to $access the value true* when $access already evaluates to true (true, 1, "some string" etc), or
whether uid values are equal and'd with the return of user_access
Correct
And otherwise assign false. After this statement $access is always either true or false, even when $access === 'yes' before.
Note*: || and && are boolean operators, only capable of 'returning' true or false
I had this exact type of statement in a library way back, and it's basically an elaborate (or maybe just badly-styled?) null-check. Because PHP uses short circuit evaluation, the right-hand side of the or-expression will not evaluate if the left hand one was null.
if ($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
....
In this snippet of code we type $inputs['user_id'] 3 times.
if (isset($inputs['user_id']) && $inputs['user_id']) { // The consumer is passing a user_id
doSomethingWith($inputs['user_id']);
}
What's the most readable and robust refactoring I can do to avoid the duplication and avoid any notice that the index user_id doesn't exist?
Thanks.
Here nothing is wrong with the duplication. You cannot assign $inputs['user_id'] to a variable before checking if it is set, otherwise this will produce a Notice undefined index ....
The only thing here could be done is to omit the isset call and use !empty instead, like this:
if(!empty($inputs['user_id'])) {
doSomething($inputs['user_id']);
}
Now You are only typing it twice and the check
!empty($inputs['user_id'])
equals to
isset($inputs['user_id']) && $inputs['user_id']
EDIT: based on a comments, here is a quote from documentation:
The following things are considered to be empty:
"" (an empty string)
0 (0 as an integer)
0.0 (0 as a float)
"0" (0 as a string)
NULL
FALSE
array() (an empty array)
$var; (a variable declared, but without a value)
So either empty(0) or empty('0') will return true, that means
if(!empty('0') || !empty(0)) { echo "SCREW YOU!"; }
will echo nothing... Or, in polite way, I will repeat the statement above:
!empty($inputs['user_id']) === (isset($inputs['user_id']) && $inputs['user_id'])
EDIT 2:
By omitting the isset and replacing by !empty the variable is still checked, whether the index is already set, please read the documentation, which says:
No warning is generated if the variable does not exist. That means empty() is essentially the concise equivalent to !isset($var) || $var == false.
What about this:
// put validation check to the function body
function doSomethingWith($userId) {
if($userId === -1) {
// if this is not a valid user id -> return
return;
}
// do something ...
}
// initalize $user with proper default values.
// doing so you can be sure that the index exists
$user = array(
'id' => -1,
'name' => '',
...
);
// merge inputs with default values:
$user = array_merge($user, $request);
// now you can just pass the value:
doSomethingWith($user['id']);
Below might not be the best way for every situation, but definitely cuts down on the repetition.
Your example code would turn into:
doSomethingWith($inputs['user_id']);
and your function would look like this (notice the argument supplied by reference, to avoid the undefined variable warning):
function doSomethingWith(&$userID) {
if (empty($userID)) return;
// ... actual code here ...
}
Assuming that 0 and "" and null are not valid user_ids:
if ($id = $inputs['user_id']) {
doer($id);
}
YOu can also do with evil # to avoid notice in your logs, (I don't like this way):
if ($id = #$inputs['user_id']) {
doer($id);
}
Possibly a strange one that I hope can be done in one line.
I have to have an IF statement that will checks two things.
The first checks if the variable $loggedInfo['status'] is set and is equal to "client".
The second checks that the variable $loggedInfo['address1'] is set and is blank.
The reason being that when the first variable equals staff then the 'address1' variable doesn't exist.
I did have the following but when I log in as staff it still checks for the address1
if((isset($loggedInfo['status'])=="client")&&(!$loggedInfo['address1'])){
//Do something
}
isset returs true or false. you have to do separate check for the actual value
if(
isset($loggedInfo['status']) && $loggedInfo['status']=="client" &&
isset($loggedInfo['address1']) && trim($loggedInfo['address1']) != ''
)
{
//Do something
}
if((isset($loggedInfo['status']) && $loggedInfo['status']=="client") &&(empty($loggedInfo['address1'])){
//Do something
}
isset() returns TRUE if the given variable is defined in the current scope with a non-null value.
empty() returns TRUE if the given variable is not defined in the current scope, or if it is defined with a value that is considered "empty". These values are:
NULL // NULL value
0 // Integer/float zero
'' // Empty string
'0' // String '0'
FALSE // Boolean FALSE
array() // empty array
Depending PHP version, an object with no properties may also be considered empty.
Well you just can't compare the return value of isset() with the string "client", because it will never equal that. To quote http://php.net/manual/en/function.isset.php its return values are "TRUE if var exists and has value other than NULL, FALSE otherwise".
First check if it is set
if ((isset($loggedInfo['status']) === true) && ($loggedInfo['status'] === "client") && (empty($loggedInfo['address1']) === true)) {
// Do something
}
Key take away from this should be to look up return values for every function you use, like empty(), in the manual http://www.php.net/manual/en/function.empty.php. This will save you a lot of headaches in the future.
I want to prevent an empty value from going into a MySQL database
I'm using following but for some reason it is letting the empty values get through...
if (trim($var)= '' || !isset($var)){
header("Location:page.php?er=novariablel");
}
else {
...insert into database
}
Note, there is a bunch of complicated stuff that sets the value of var which is why I want to have both the ='' and the !isset because either might be the case.
Am I missing something with the or statement, i.e. it evaluates to false if both are true. Or what am I doing wrong?
You're lacking an = for your Equal Comparison Operator inside your if statement. Try:
if (trim($var) == '' || !isset($var)){
Try:
if (!isset($var) || empty(trim($var))){
empty() is a better way to check to see if a variable has no value. Just keep in mind that the following will return true:
The following things are considered to be empty:
"" (an empty string)
0 (0 as an integer)
0.0 (0 as a float)
"0" (0 as a string)
NULL
FALSE
array() (an empty array)
var $var; (a variable declared, but without a value in a class)
Try:
$var = trim($var);
if (!empty($var)){
//not empty
}
Try
if (strlen(trim($var)))==0
You should also have put a constraint in your database table attribute that it will not accept NULL values. Then your entry would have not been made in the database.
#John this is great advice.
empty(trim($var))
I've been programming for 6 years and I never thought of trimming the variable before checking to see if it's empty.