Say I got some function that run some code and then return something, like this:
function something()
{
//some code
return $some[$whatever];
}
So, if I want to extract the data I generated in the function - the new value for $some, how should I do it? for example this won't do anything:
echo ($some);
Or what am I missing here, please
Since your Function returns a value, You may need to catch & store it inside a variable and then echo the variable if it is a String or do some casting to that effect. Here's an example:
<?php
function something(){
//some code
$whatever = 3;
$some = ["Peace", "Amongst", "All", "Humanity"];
return $some[$whatever];
}
$var = something();
var_dump($var); //<== DUMPS :: "Humanity"
echo $var; //<== ECHOES:: "Humanity"
Test it out here.
Cheers and Good Luck....
You are trying to return a specif key from your array, which wasn't declared. I declared an array for you, and I added the isset to check if the key is existing in the array to prevent any php warnings.
function something($findKey)
{
$some = array('key'=> 123);
if(!isset($some[$findKey])) {
return false;
}
//some code
return $some[$findKey];
}
echo something('key');
Related
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 $sizeShown variable, you should have what you want.
Iwonder why the out put is 0(zero) for below code snippet? Can anyone please clarify why below code output is zero.
<?php
function a($number)
{
return (b($number) * $number);
}
function b(&$number)
{
++$number;
}
echo a(5); // output 0(zero) ?
?>
You never return any value from the function, and you're trying to echo the return value.
function b(&$number)
{
return ++$number;
}
Note that this is a silly example for a function that takes its parameter by reference, since you don't have a reference to the original value 5. Something like this would be more appropriate:
function b( &$number) {
++$number;
}
$num = 5;
b( $num);
echo $num; // Prints 6
The function name is b, but you are calling a...
Also, you are echoing a function, that doesn't return a value. This means you are echoing a non-initialize variable.
You must either return a value:
return ++$number;
or echo the variable directly:
$number = 5;
b($number);
echo $number;
I am having trouble returning a value from a multidimensional array using a key ONLY when I externalize to create a function.
To be specific, the following code will work when inline in a page:
<?php
foreach ($uiStringArray as $key) {
$keyVal = $key['uid'];
if($keyVal == 'global001') echo $key['uiString'];
}
?>
However, if I externalize the code as a function, like so:
function getUIString($myKey) {
// step through the string array and find the key that matches the uid,
// then return uiString
$myString = "-1";
foreach ($uiStringArray as $key) {
$keyVal = $key['uid'];
if($keyVal == 'global001') {
$myString = $key['uiString'];
}
}
return $myString;
}
And then call it like this:
<?php getUIString('global001'); ?>
It always returns -1, and will do so even if I use an explicit key in the function rather than a variable. I can't understand why this works inline, but fails as a function.
I'm a relative PHP noob, so please forgive me if this includes a glaring error on my part, but I've searched all over for discussion of this behavior and found none.
All help appreciated.
i think you need to take a look at PHP's Variable Scope. To problem is that PHP isn't typical of other languages where a variable defined outside of a function is visible within. You need to use something like the $GLOBALS variable or declare the variable global to access it.
To better illustrate, picture the following:
$foo = "bar";
function a(){
// $foo is not visible
echo $foo;
}
function b(){
global $foo; // make $foo visible
echo $foo;
}
function c(){
// acccess foo within the global space
echo $GLOBALS['foo'];
}
The same is basically holding true for your $uiStringArray variable in this scenario.
This is a problem with variable scope, see Brad Christie's answer for more details on variable scope.
As for your example, you need to either pass the array to the function or create it inside the function. Try:
function getUIString($myKey, $uiStringArray = array()) {
// step through the string array and find the key that matches the uid,
// then return uiString
$myString = "-1";
foreach ($uiStringArray as $key) {
$keyVal = $key['uid'];
if($keyVal == 'global001') {
$myString = $key['uiString'];
break;
}
}
return $myString;
}
And call the function using
<?php getUIString('global001', $uiStringArray); ?>
You are having this problem because you are overriding you $mystring variable even if it matches. Send your array as your parameter. It is unknown to your function.You just use break if variable matches
function getUIString($myKey, $uiStringArray=array()) {
// step through the string array and find the key that matches the uid,
// then return uiString
$myString = "-1";
foreach ($uiStringArray as $key) {
$keyVal = $key['uid'];
if($keyVal == 'global001') {
$myString = $key['uiString'];
break;
}
}
return $myString;
}
so I have 2 functions like this:
function boo(){
return "boo";
}
and
function foo(){
echo "foo";
}
the fist one will return a value, and the 2nd one will output something to the screen directly.
$var = boo();
foo();
How can I merge these 2 functions into one, and somehow detect if it's being called to output the result to the screen, or if it's called for getting the return value? Then choose to use return or echo...
function boo_or_foo ($output = false) {
if ($output) {
echo "fbo";
} else {
return "foo";
}
}
But whats the benefit against just using one function (boo()) and echo it yourself?
echo $boo();
Well, a function should only do one thing, so typically you would have two functions. But, if you would like to combine them you can just check if is set:
function boo($var=null){
if(isset($var)) echo $var
else return "boo";
}
well return true in the function that prints then yo just do
function foo(){
echo "foo";
return true;
}
if(foo()){
echo "foo did print something";
}else{
echo "nope foo is broken";
}
I wanted to achieve the same effect. In my case I have functions that produce HTML which I want echoed directly sometimes (when an Ajax call is being made), or returned (when a call is made from another script).
For example, a function that creates a list of HTML <option> elements - listOfOption($filter). When one of my pages is first created, the function is called and the result is echoed in place:
<?= listOfOption($var) ?>
But sometimes the same data needs to be retrieved in an Ajax call:
http://site.com/listOfOption.php?parameter=2
Instead of writing two different scripts or specifying the behaviour in a parameter, I keep listOfOption($filter) in its own file like this:
if (__FILE__ == $_SERVER['SCRIPT_FILENAME'])
{
echo listOfOption($_REQUEST['parameter']);
}
function listOfOption($filter)
{
return '<option value="1">Foo</option>';
}
This way if the call is from another script, it returns the data; otherwise it prints the data.
Note that if a parameter isn't passed to the function I wouldn't have to do this, I could live with echoing the data always and replacing the <?= listOfOption() ?> invocation with <? listOfOption() ?> to keep things clear.
What does the & before the function name signify?
Does that mean that the $result is returned by reference rather than by value?
If yes then is it correct? As I remember you cannot return a reference to a local variable as it vanishes once the function exits.
function &query($sql) {
// ...
$result = mysql_query($sql);
return $result;
}
Also where does such a syntax get used in practice ?
Does that mean that the $result is returned by reference rather than by value?
Yes.
Also where does such a syntax get used in practice ?
This is more prevalent in PHP 4 scripts where objects were passed around by value by default.
To answer the second part of your question, here a place there I had to use it: Magic getters!
class FooBar {
private $properties = array();
public function &__get($name) {
return $this->properties[$name];
}
public function __set($name, $value) {
$this->properties[$name] = $value;
}
}
If I hadn't used & there, this wouldn't be possible:
$foobar = new FooBar;
$foobar->subArray = array();
$foobar->subArray['FooBar'] = 'Hallo World!';
Instead PHP would thrown an error saying something like 'cannot indirectly modify overloaded property'.
Okay, this is probably only a hack to get round some maldesign in PHP, but it's still useful.
But honestly, I can't think right now of another example. But I bet there are some rare use cases...
Does that mean that the $result is returned by reference rather than by value?
No. The difference is that it can be returned by reference. For instance:
<?php
function &a(&$c) {
return $c;
}
$c = 1;
$d = a($c);
$d++;
echo $c; //echoes 1, not 2!
To return by reference you'd have to do:
<?php
function &a(&$c) {
return $c;
}
$c = 1;
$d = &a($c);
$d++;
echo $c; //echoes 2
Also where does such a syntax get used in practice ?
In practice, you use whenever you want the caller of your function to manipulate data that is owned by the callee without telling him. This is rarely used because it's a violation of encapsulation – you could set the returned reference to any value you want; the callee won't be able to validate it.
nikic gives a great example of when this is used in practice.
<?php
// You may have wondered how a PHP function defined as below behaves:
function &config_byref()
{
static $var = "hello";
return $var;
}
// the value we get is "hello"
$byref_initial = config_byref();
// let's change the value
$byref_initial = "world";
// Let's get the value again and see
echo "Byref, new value: " . config_byref() . "\n"; // We still get "hello"
// However, let’s make a small change:
// We’ve added an ampersand to the function call as well. In this case, the function returns "world", which is the new value.
// the value we get is "hello"
$byref_initial = &config_byref();
// let's change the value
$byref_initial = "world";
// Let's get the value again and see
echo "Byref, new value: " . config_byref() . "\n"; // We now get "world"
// If you define the function without the ampersand, like follows:
// function config_byref()
// {
// static $var = "hello";
// return $var;
// }
// Then both the test cases that we had previously would return "hello", regardless of whether you put ampersand in the function call or not.