In codeigniter what is the correct method of passing variables to the callback function?
I have used this,
$var1 = 'some conditions';
$this->form_validation->set_rules("callback__is_value_unique[value, $var1]");
public function _is_value_unique($value, $var1){
echo $var1;
die;
}
This gave me output like shown below:-
value, some conditions
rather than,
some conditions
You must only set your value!
$var1 = 'some conditions';
$this->form_validation->set_rules("callback__is_value_unique[value, $var1]");
public function _is_value_unique($value, $var1){
echo $var1;
die;
}
From the docs..
If you need to receive an extra parameter in your callback function,
just add it normally after the function name between square brackets,
as in: "callback_foo[bar]", then it will be passed as the second
argument of your callback function.
Sounds like you could only pass one extra argument. Which should be a string. If you want to pass more arguments, you could store them somewhere else and just pass an argument, which stores the location of the additional param.
$index = count($this->arguments);
$this->arguments[$index] = array('value', 'some conditions'/*, ...*/);
$this->form_validation->set_rules("callback__is_value_unique[$index]");
public function _is_value_unique($value, $index){
$args = $this->argumts[$index];
echo $args[1];
die;
}
Related
I'm trying to write a simple function which takes two arguments, adds them together and returns the result of the calculation.
Before performing the calculation the function checks whether either of the two arguments are undefined and if so, sets the argument to 0.
Here's my function:
Function - PHP
function returnZeroAdd ($arg, $arg2)
{
if(!isset($arg))
{
$arg = 0;
}
if(!isset($arg2))
{
$arg2 = 0;
}
echo $arg + $arg2;
}
I've tried to execute it like so :
returnZeroAdd($bawtryReturnCount, $bawtryFReturnCount);
But this throws up an undefined variable $bawtryFReturnCount error.
I do not know why the function isn't setting $bawtryFReturnCount) to 0 before performing the calculation thereby negating the 'undefined variable' error.
Can anybody provide a solution?
You cannot do this the way you want. As soon as you use an undefined variable, you will get this error. So the error doesn't occur inside your function, but already occurs in the call to your function.
1. Optional parameters
You might make a parameter optional, like so:
function returnZeroAdd ($arg = 0, $arg2 = 0)
{
return $arg + $arg2;
}
This way, the parameter is optional, and you can call the function like this:
echo returnZeroAdd(); // 0
echo returnZeroAdd(1); // 1
echo returnZeroAdd(1, 1); // 2
2. By reference
But I'm not sure if that is what you want. This call will still fail:
echo returnZeroAdd($undefinedVariable);
That can be solved by passing the variables by reference. You can then check if the values are set and if so, use them in the addition.
<?php
function returnZeroAdd (&$arg, &$arg2)
{
$result = 0;
if(isset($arg))
{
$result += $arg;
}
if(isset($arg2))
{
$result += $arg2;
}
return $result;
}
echo returnZeroAdd($x, $y);
Note that you will actually change the original value of a by reference parameter, if you change it in the function. That's why I changed the code in such a way that the parameters themselves are not modified. Look at this simplified example to see what I mean:
<?php
function example(&$arg)
{
if(!isset($arg))
{
$arg = 0;
}
return $arg;
}
echo example($x); // 0
echo $x // also 0
Of course that might be your intention. If so, you can safely set $arg and $arg2 to 0 inside the function.
The error is not thrown by the function itself, as the function is not aware of the global scope. The error is thrown before even the function is executed, while the PHP interperter is trying to pass $bawtryFReturnCount to the function, one does not find it, and throws error, however, it's not a fatal one and the execution is not stopped. THerefore, the function is executed with a non-set variable with default value of null, where I guess, isset will not work, as the arguments are mandatory, but not optional. A better check here will be empty($arg), however the error will still be present.
Because the functions are not and SHOULD NOT be aware of the global state of your application, you should do these checks from outside the functions and then call it.
if (!isset($bawtryReturnCount)) {
$bawtryReturnCount = 0
}
returnZeroAdd($bawtryReturnCount);
Or assign default values to the arguments in the function, making them optional instead of mandatory.
Your function could be rewritten as:
function returnZeroAdd ($arg = 0, $arg2 = 0)
{
echo $arg + $arg2;
}
You missunderstand how variables work. Since $bawtryFReturnCount isn't defined when you call the function; you get a warning. Your isset-checks performs the checks too late. Example:
$bawtryReturnCount = 4;
$bawtryFReturnCount = 0;
returnZeroAdd($bawtryReturnCount, $bawtryFReturnCount);
Will not result in an error.
If you really want to make the check inside the function you could pass the arguments by reference:
function returnZeroAdd (&$arg, &$arg2)
{
if(!isset($arg))
{
$arg = 0;
}
if(!isset($arg2))
{
$arg2 = 0;
}
echo $arg + $arg2;
}
However this will potentially modify your arguments outside the function, if it is not what you intend to do then you need this:
function returnZeroAdd (&$arg, &$arg2)
{
if(!isset($arg))
{
$localArg = 0;
}
else
{
$localArg = $arg;
}
if(!isset($arg2))
{
$localArg2 = 0;
}
else
{
$localArg2 = $arg2;
}
echo $localArg + $localArg2;
}
You can now pass undefined variables, it won't throw any error.
Alternatively you might want to give a default value to your arguments (in your case 0 seems appropriate):
function returnZeroAdd ($arg = 0, $arg2 = 0)
{
echo $arg + $arg2;
}
You have to define the variable before pass it to an function. for example
$bawtryReturnCount=10;
$bawtryFReturnCount=5;
define the two variable with some value and pass it to that function.
function returnZeroAdd ($arg=0, $arg2=0)
{
echo $arg + $arg2;
}
if you define a function like this means the function takes default value as 0 if the argument is not passed.
for example you can call the functio like this
returnZeroadd();// print 0
returnZeroadd(4);// print 4
returnZeroadd(4,5);// print 9
or you can define two variables and pass it as an argument and call like this.
$bawtryReturnCount=10;
$bawtryFReturnCount=5;
returnZeroadd($bawtryReturnCount, $bawtryFReturnCount);
How would I alter the function below to produce a new variable for use outside of the function?
PHP Function
function sizeShown ($size)
{
// *** Continental Adult Sizes ***
if (strpos($size, 'continental-')!== false)
{
$size = preg_replace("/\D+/", '', $size);
$searchsize = 'quantity_c_size_' . $size;
}
return $searchsize;
Example
<?php
sizeShown($size);
$searchsize;
?>
This currently produces a null value and Notice: undefined variable.
So the function takes one argument, a variable containing a string relating to size. It checks the variable for the string 'continental-', if found it trims the string of everything except the numbers. A new variable $searchsize is created which appends 'quantity_c_size_' to the value stored in $size.
So the result would be like so ... quantity_c_size_45
I want to be able to call $searchsize outside of the function within the same script.
Can anybody provide a solution?
Thanks.
Try using the global keyword, like so:
function test () {
global $test_var;
$test_var = 'Hello World!';
}
test();
echo $test_var;
However, this is usually not a good coding practice. So I would suggest the following:
function test () {
return 'Hello World!';
}
$test_var = test();
echo $test_var;
In the function 'sizeShown' you are just returning the function. You forgot to echo the function when you call your function.
echo sizeShown($size);
echo $searchsize;
?>
But the way you call $searchsize is not possible.
This is an old question, and I might not be understanding the OP's question properly, but why couldn't you just do this:
<?php
$searchsize = sizeShown($size);
?>
You're already returning $searchsize from the sizeShown method. So if you simply assign the result of the function to the $sizeShown variable, you should have what you want.
I am playing around with this:
$sort = array('t1','t2');
function test($e){
echo array_search($e,$sort);
}
test('t1');
and get this error:
Warning: array_search(): Wrong datatype for second argument on line 4
if I call it without function like this, I got the result 0;
echo array_search('t1',$sort);
What goes wrong here?? thanks for help.
Variables in PHP have function scope. The variable $sort is not available in your function test, because you have not passed it in. You'll have to pass it into the function as a parameter as well, or define it inside the function.
You can also use the global keyword, but it is really not recommended. Pass data explictly.
You must pass the array as a parameter! Because the functions variables are different from globals in php!
Here is the fixed one:
$sort = array('t1','t2');
function test($e,$sort){
echo array_search($e,$sort);
}
test('t2',$sort);
You cannot directly access global variables from inside functions.
You have three options:
function test($e) {
global $sort;
echo array_search($e, $sort);
}
function test($e) {
echo array_search($e, $GLOBALS['sort']);
}
function test($e, $sort) {
echo array_search($e, $sort);
} // call with test('t1', $sort);
take the $sort inside the function or pass $sort as parameter to function test()..
For e.g.
function test($e){
$sort = array('t1','t2');
echo array_search($e,$sort);
}
test('t1');
----- OR -----
$sort = array('t1','t2');
function test($e,$sort){
echo array_search($e,$sort);
}
test('t1',$sort);
I have a function variable like this...
$aName = "My Name";
The function need to pass in like this:
$sayHelloFunction = function sayHello($aName){
echo($aName);
}
Than, I have a method that responsible for execute the sayHelloFunction:
runningFunction($sayHelloFunction($aName));
in the runningFunction, it will have some condition to execute the function, but when I pass the "$sayHelloFunction($aName)" to runningFunction, it execute automatically, but I would like to pass the variable $aName as well, how can I achieve it? Thank you.
runningFunction($sayHelloFunction, $aName);
Simples.
You will have to pass the arguments separately. However, you could wrap them in an array so that you can pass them to runningFunction as a single argument, like this:
$printFunction = function($args) {
print $args['lastname'].', '.$args['firstname'];
};
function runningFunction($f, $a) {
$f($a);
}
$firstname = 'Bob';
$lastname = 'Smith';
$functionArguments = array(
'firstname' => $firstname,
'lastname' => $lastname
);
runningFunction($printFunction, $functionArguments);
If you want your dynamic functions to get "proper" arguments, then I see no way around something like this:
function runningFunction($f, $a) {
switch(count($a)) {
0: $f(); break;
1: $f($a[0]); break;
2: $f($a[0], $a[1]); break;
3: $f($a[0], $a[1], $a[2]); break;
// and so on
}
}
Pass the parameters as an array, and then use call_user_func_array() to call your function.
This way your runningFunction() will be absolutely abstract (as you requested), it can call any type of function, it's your responsibility to pass the right number of parameters.
function runningFunction ($callback, $parameters=array()) {
call_user_func_array($callback, $parameters);
}
runningFunction($sayHelloFunction, array($aName));
call_user_func_array()
as xconspirisist suggested pass $aName as a seperate parameter to the function.
Details on Variable Functions can be found on the PHP site.
Use an anonymous function when calling runningFunction
function runningFunction($func) {
$func();
}
runningFunction(function() {
$sayHelloFunction($aName));
// You can do more function calls here...
});
I have a function to send mail to users and I want to pass one of its parameter as an array of ids.
Is this possible to do? If yes, how can it be done?
Suppose we have a function as:
function sendemail($id, $userid) {
}
In the example, $id should be an array.
You can pass an array as an argument. It is copied by value (or COW'd, which essentially means the same to you), so you can array_pop() (and similar) all you like on it and won't affect anything outside.
function sendemail($id, $userid){
// ...
}
sendemail(array('a', 'b', 'c'), 10);
You can in fact only accept an array there by placing its type in the function's argument signature...
function sendemail(array $id, $userid){
// ...
}
You can also call the function with its arguments as an array...
call_user_func_array('sendemail', array('argument1', 'argument2'));
even more cool, you can pass a variable count of parameters to a function like this:
function sendmail(...$users){
foreach($users as $user){
}
}
sendmail('user1','user2','user3');
Yes, you can safely pass an array as a parameter.
Yes, you can do that.
function sendemail($id_list,$userid){
foreach($id_list as $id) {
printf("$id\n"); // Will run twice, once outputting id1, then id2
}
}
$idl = Array("id1", "id2");
$uid = "userID";
sendemail($idl, $uid);
What should be clarified here.
Just pass the array when you call this function.
function sendemail($id,$userid){
Some Process....
}
$id=array(1,2);
sendmail($id,$userid);
function sendemail(Array $id,$userid){ // forces $id must be an array
Some Process....
}
$ids = array(121,122,123);
sendmail($ids, $userId);
Its no different to any other variable, e.g.
function sendemail($id,$userid){
echo $arr["foo"];
}
$arr = array("foo" => "bar");
sendemail($arr, $userid);
I composed this code as an example. Hope the idea works!
<?php
$friends = array('Robert', 'Louis', 'Ferdinand');
function greetings($friends){
echo "Greetings, $friends <br>";
}
foreach ($friends as $friend) {
greetings($friend);
}
?>
I found Delcon answer helpful but I was looking for this
function sendmail($user1, $user2, $user3){
echo $user1;
echo $user2;
echo $user3;
}
$users = array('user1','user2','user3');
sendmail(...$users);
In php 5, you can also hint the type of the passed variable:
function sendemail(array $id, $userid){
//function body
}
See type hinting.
Since PHP is dynamically weakly typed, you can pass any variable to the function and the function will try to do its best with it.
Therefore, you can indeed pass arrays as parameters.
Yes, we can pass arrays to a function.
$arr = array(“a” => “first”, “b” => “second”, “c” => “third”);
function user_defined($item, $key)
{
echo $key.”-”.$item.”<br/>”;
}
array_walk($arr, ‘user_defined’);
We can find more array functions here
http://skillrow.com/array-functions-in-php-part1/
<?php
function takes_array($input)
{
echo "$input[0] + $input[1] = ", $input[0]+$input[1];
}
?>