I have a function which return a loop, now I want to check if this function returns null or empty
That's the function:
function getCopy($pname){
function listCopy($block) {
foreach ($block as $b) {
echo '<div class="copy">'.$b->getBlock().'</div>'. "\n";
}
}
$filter=array(
new DFC(classContent::FIELD_PNAME, $pname, DFC::CONTAINS),
new DFC(classContent::FIELD_TYPE, 'copy', DFC::CONTAINS),
);
$block=classContent::findByFilter(conn(), $filter);
return listCopy($block);
}
That's my logic:
if( isset (getCopy($pname)) ){
echo "<label for='copy'>Copy</label><br>"
."<textarea name='copy' id='copy' rows='10' cols='60'>".getCopy($pname)
."</textarea><br>";
}
The isset doesn't work and neither if(getCopy($pname) != '') does.
Any idea on how to do this?
Thanks in advance
Mauro
use empty or is_null or a combination of both.
Or you could just negate the check by doing if( !getCopy($pname) ) ){ ... } but i'd go with any of the two functions above.
Edit: as deceze noted, you can't directly evaluate the return value with empty, you'll have to assign it to a var first and then pass that var to empty()
$result = getCopy($pname);
if(empty($result)) { ... }
You can't "return a loop" from a function. You can only return values. The listCopy function does not return anything, it just outputs. Hence getCopy doesn't return anything either. Defining a function within a function is usually bad practice as well, you won't be able to call getCopy twice in your case.
I'm not really sure what you're trying to do, but you need to rethink your approach.
The php function isset()is used to determine if a variable is set and is not NULL. So it's normal that it is not working.
empty() is what you are looking for but bare in mind these two points:
You shouldn't declare functions within functions.
Your function will never return something, it's always void since you don't return anything. So empty() will always evaluate to true.
Related
I am confused about return statement , why we need to use return in end of function , for example
function test($a){blah;
blahh ;
blahhh;
blahhhhh;
return;}
What the use of return here? Function automatically terminates when all the statements executed , I think there is no use of return here , but this picture from http://www.w3resource.com/php/statement/return.php make me confused
So
Can someone please explain the use of return (when we not returning any value).
It depends on what you're trying to achieve.
If you write echo in several places, your code will get confusing. In general, a function that returns a value is also more versatile, since the caller can decide whether to further manipulate that value or immediately print it.
I'd recommend to stick to the convention and use return for a function.
You should check GuardClause.
Example:
function test() {
return 10;
}
$a = test(); // $a stores the value 10
echo $a; // Prints 10
echo $a + 5; // We may want to manipulate the value returned by the function. So, it prints 15.
For further reference, check What is the difference between PHP echo and PHP return in plain English?
In that context: You don't.
return breaks out of the function, but since it is the last statement in that function, you would break out of it anyway without the statement.
return passes its argument back to the caller, but since it doesn't have an argument, there is nothing to pass.
So it does nothing.
Don't assume that every piece of code you stumble across has a purpuse. It might be left over from an earlier version of the code where something else (which gave it meaning) has been removed. It might be written by someone cargo culting. It might be placeholder for future development.
"return" is important when you plan to call this function from other codes, it helps you to:
Know if the function works correctly, or not.
Obtain values from a function.
Make sure other codes are not executed after return.
It might be useless when the code is simple as your sample, let's make it more complex.
function test($a){
if(file_exists($a)){
if(is_file($a)){
return $a." IS A FILE\n";
} else if(is_dir($a)) {
return $a." IS A DIR\n";
} else {
return $a." EXISTS, BUT I DONT KNOW WHAT IT IS\n"
}
} else {
return $a." NOT EXISTS\n";
}
return 0;
}
$filecheck = test("/abc/def.txt");
if($filecheck){
echo $filecheck;
} else {
echo "unknown error";
}
Above shows how to return a value, and do some basic handling.
It is always good to implement return in your functions so you can even specify error code for your functions for reference.
Based on your example, I'll modify slightly like below:
function test($a){
blah;
blahh;
blahhh;
blahhhhh;
return 1;
}
So I know the code is running to last line.
Otherwise the function just finishes silently.
It's useful if you want a function to return a value.
i.e.
Function FavouritePie($who) {
switch($who) {
case 'John':
return 'apple';
case 'Peter':
return 'Rhubarb';
}
}
So, considering the following:
$WhatPie = FavouritePie('John');
echo $WhatPie;
would give
apple
In a really simple form, it's useful if you want a function to return something, i.e. process it and pass something back. Without a return, you'd just be performing a function with a dead end.
Some further reading:
http://php.net/manual/en/function.return.php
http://php.net/manual/en/functions.returning-values.php
To add some further context specific to the answer, if the question boils down to "Do I need to add a return for the sake of it, at the end of a function, then the answer is no. But that doesn't mean the correct answer is always to leave out your return.
it depends what you want to do with $a.
If you echo $a within the function, that function will spit out $a as soon as it is called. If you return $a, assuming you set the calling of test to a variable (i.e. $something = $test('foo')), then you can use it later on.
So, I am writing a function that receives two arguments, and before I do anything with these variables, I want to check if they are not null.
if (!is_null($foo) && !is_null($bar)) {
/* do something */
}
The problem is that I think we are repeating code, and in some cases when the variable name is a little bit bigger, it becomes painful to write every time.
So, there is a way to shorten this code?
You can use isset which returns FALSE if variable is null. You can set there a lot of variables so code will be shorter:
if (isset($foo, $bar))
You can write your own function to check this:
function is_any_null() {
$params = func_get_args();
foreach($params as $param) {
if (is_null($param))
return true;
}
return false;
}
Now you can use it like this:
if (!is_any_null($foo, $bar)) {
/* do something */
}
is_null() checks a variable to determine if the value is NULL. It can be used in an if statement. It depends on what you would like to do in that if statement. For example, if you would like to do an echo, you can make it a bit shorter using the `elvis``operator:
echo (is_null($foo) && is_null($bar)) ?: 'The values are not null';
Or you could make it a one liner by leaving the { and } out of your code and putting everything on one line.
if (!is_null($foo) && !is_null($bar)) die("NULLLLL");
I have an interesting situation. I am using a form that is included on multiple pages (for simplicity and to reduce duplication) and this form in some areas is populated with values from a DB. However, not all of these values will always be present. For instance, I could be doing something to the effect of:
<?php echo set_value('first_name', $first_name); ?>
and this would work fine where the values exist, but $user is not always set, since they may be typing their name in for the first time. Yes you can do isset($first_name) && $first_name inside an if statement (shorthand or regular)
I am trying to write a helper function to check if a variable isset and if it's not null. I would ideally like to do something like varIsset('first_name'), where first_name is an actual variable name $first_name and the function would take in the string, turn it into the intended variable $first_name and check if it's set and not null. If it passes the requirements, then return that variables value (in this case 'test'). If it doesn't pass the requirements, meaining it's not set or is null, then the function would return '{blank}'.
I am using CodeIgniter if that helps, will be switching to Laravel in the somewhat near future. Any help is appreciated. Here is what I've put together so far, but to no avail.
function varIsset($var = '')
{
foreach (get_defined_vars() as $val) {
if ($val == $var) {
if (isset($val) && $val) {
echo $val;
}
break;
}
}
die;
}
Here is an example usage:
<?php
if (varIsset('user_id') == 100) {
// do something
}
?>
I would use arrays and check for array keys myself (or initialize all my variables...), but for your function you could use something like:
function varIsset($var)
{
global $$var;
return isset($$var) && !empty($$var);
}
Check out the manual on variable variables. You need to use global $$var; to get around the scope problem, so it's a bit of a nasty solution. See a working example here.
Edit: If you need the value returned, you could do something like:
function valueVar($var)
{
global $$var;
return (isset($$var) && !empty($$var)) ? $$var : NULL;
}
But to be honest, using variables like that when they might or might not exist seems a bit wrong to me.
It would be a better approach to introduce a context in which you want to search, e.g.:
function varIsset($name, array $context)
{
return !empty($context[$name]);
}
The context is then populated with your database results before rendering takes place. Btw, empty() has a small caveat with the string value "0"; in those cases it might be a better approach to use this logic:
return isset($context[$name]) && strlen($name);
Try:
<?php
function varIsset($string){
global $$string;
return empty($$string) ? 0 : 1;
}
$what = 'good';
echo 'what:'.varIsset('what').'; now:'.varIsset('now');
?>
I have found there to be multiple ways to check whether a function has correctly returned a value to the variable, for example:
Example I
$somevariable = '';
$somevariable = get_somevariable();
if ($somevariable)
{
// Do something because $somevariable is definitely not null or empty!
}
Example II
$somevariable = '';
$somevariable = get_somevariable();
if ($somevariable <> '')
{
// Do something because $somevariable is definitely not null or empty!
}
My question: what is the best practice for checking whether a variable is correct or not? Could it be different for different types of objects? For instance, if you are expecting $somevariable to be a number, would checking if it is an empty string help/post issues? What is you were to set $somevariable = 0; as its initial value?
I come from the strongly-typed world of C# so I am still trying to wrap my head around all of this.
William
It depends what you are looking for.
Check that the Variable is set:
if (isset($var))
{
echo "Var is set";
}
Checking for a number:
if (is_int($var))
{
echo "Var is a number";
}
Checking for a string:
if (is_string($var))
{
echo "var is a string";
}
Check if var contains a decimal place:
if (is_float($var))
{
echo "Var is float";
}
if you are wanting to check that the variable is not a certain type, Add: ! an exclamation mark. Example:
if (!isset($var)) // If variable is not set
{
echo "Var Is Not Set";
}
References:
http://www.php.net/manual/en/function.is-int.php
http://www.php.net/manual/en/function.is-string.php
http://www.php.net/manual/en/function.is-float.php
http://www.php.net/manual/en/function.isset.php
There is no definite answer since it depends on what the function is supposed to return, if properly documented.
For example, if the function fails by returning null, you can check using if (!is_null($retval)).
If the function fails by returning FALSE, use if ($retval !== FALSE).
If the function fails by not returning an integer value, if (is_int($retval)).
If the function fails by returning an empty string, you can use if (!empty($retval)).
and so on...
It depends on what your function may return. This kind of goes back to how to best structure functions. You should learn the PHP truth tables once and apply them. All the following things as considered falsey:
'' (empty string)
0
0.0
'0'
null
false
array() (empty array)
Everything else is truthy. If your function returns one of the above as "failed" return code and anything else as success, the most idiomatic check is:
if (!$value)
If the function may return both 0 and false (like strpos does, for example), you need to apply a more rigorous check:
if (strpos('foo', 'bar') !== false)
I'd always go with the shortest, most readable version that is not prone to false positives, which is typically if ($var)/if (!$var).
If you want to check whether is a number or not, you should make use of filter functions.
For example:
if (!filter_var($_GET['num'], FILTER_VALIDATE_INT)){
//not a number
}
I have the following PHP function
function newUser($username, $password) {
$checkUsernameAvailablity = check($username);
if (!$checkUsernameAvailablity)
{
return -1;
}
$checkPasswordComplexity = checkpass($password);
if (!$checkPasswordComplexity)
{
return -2
}
}
I would like to know if the username is taken and the password is not complex enough, will PHP stop the function after it returns -1 or will it continue and return -2 also.
Thanks in advance,
RayQuang
When execution reaches a return statement, the function will stop and return that value without processing any more of the function.
return statement returns the value and terminates the execution of the function. If the return is hit, no further code is executed in that body.
By the way, not all code paths have return values.
the design of your function is wrong.
I would return different values when the function ends accordingly.
For multiple returns I would use switch or a properly designed IF statements.
Or better return an associative array.