I got strange problem, when i doing echo $setting->samp->url; it works perfectly but when i put it inside a function. It show me Notice undefined variable, what happened here??
$setting = simplexml_load_file('setting.xml');
#Content inside <samp><url>value</url></samp> printed.
echo $setting->samp->url;
sampserver_update();
#Problem
function sampserver_update() {
echo $setting->samp->url;
}
That's because $setting is not available inside the function. You might want to pass it as an argument.
Read about variable scopes here:
http://php.net/manual/en/language.variables.scope.php
Value must be passed as argument to function
sampserver_update($setting);
function sampserver_update($setting) {
echo $setting->samp->url;
}
just replace this lines
Update your function to following
function sampserver_update()
{
$setting = simplexml_load_file('setting.xml');
return $setting->samp->url;
}
Calling and print
echo sampserver_update();
Related
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');
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);
Can anyone tell me why this works (it echos out "poo"):
$input = "wee";
$val = "poo";
${$input} = $val;
echo $wee;
But this doesn't:
function bodily($input) {
$val = "poo";
${$input} = $val;
}
bodily("wee");
echo $wee;
I want to use this sort of method to play with some $_POST vars. Please tell me if I can explain more... Cheers!
Your variable $wee gets only defined inside the scope of your function bodily(). It is not defined outside this function.
You could make it global, anyway this is not a useful pattern for a real life application:
function bodily($input) {
$val = "poo";
global ${$input}; // make your $wee defined in the global scope
${$input} = $val;
}
bodily("wee");
echo $wee;
outputs
poo
Because the variable is defined locally inside of the function. Let the function return the value and assign it to a variable outside of the function.
Because the variables inside a function are not accessible from outside unless inside the function you use "global $var" or pass it by reference like function (&$var) ...
in order for your code to work you need
<?php
function bodily($input) {
$val = "poo";
${$input} = $val;
echo $wee;
}
bodily("wee");
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.