Running a loop that has a function inside it - php

Question

I have a problem trying to run a code inside the loop, my loop consist of a function.
Here is my coding:
$new = array(1,2,3,4);
for($i=0;$i<=3;$i++){
$val = $new[$i];
function myfunction($value) {
//Do something
}
echo $val;
}
The problem is the code outputs only the 1st value in my array. I am very confused, am I not suppose to declare a function inside the loop?

Answer 1

Your code ends up with Fatal error, since at the second iteration it tries to redeclare function myfunction. That's why it is printing only first value of array.
In order to avoid that fatal error you can check if that function has been already defined using function_exists() function like this:
$new = array(1,2,3,4);
for($i=0;$i<=3;$i++)
{
$val = $new[$i];
if(!function_exists('myfunction'))
{
function myfunction($value) {
//Do something
}
}
echo $val;
}
PHP is a scripting language and it is syntactically correct to declare a function inside for loop or inside if statement, but it is a bad practice and can cause a lot of errors afterwards.
The best way is to declare a function outside loop, and, if needed, call it from within a loop like this:
<?php
function myfunction($value) {
//Do something
}
$new = array(1,2,3,4);
for($i=0;$i<=3;$i++)
{
$val = $new[$i];
myfunction($value); //may you was intended to pass $val here?
echo $val;
}

Answer 2

Don't declare the function inside the loop, declare it before the loop and then call to it inside the loop with myFunction($value);

Answer 3

the function should be in a separate procedure
$new = array(1,2,3,4);
for($i=0;$i<=3;$i++)
{
$val = $new[$i];
myfunction($val)
echo $val;
}
then this is your function
function myfunction($value)
{
//Do something
}

Answer 4

Declare the function outside of the loop
either return a value from the function, or let the function output data
For example:
function myfunction($value) {
//Do something
echo $value;
}
$new = array(1,2,3,4);
for($i=0;$i<=3;$i++) {
myfunction($new[$i]);
}

Answer 5

I am assuming you want to print out the first 4 elements of the array.
do something like this
function myfunction() {
$new = array(1,2,3,4);
for($i=0;$i<=3;$i++){
$val = $new[$i];
echo $val;
}
}
myfunction();

Answer 6

You should declare the function outside the loop
function myfunction($value) {
return ($value + 25); // an example
}
$new = array(1,2,3,4);
for($i = 0; $i < count($new); $i++){
echo myfunction($new[$i]);
}
Also you should set the loop from 0 to the end of the array, so if you'll have more than 4 entries in the array the code should be ok

Answer 7

You can declare an anonymous function instead:
for ($i=0; $i<=3; $i++) {
// code
$myFunction = function($value) { /* code */ }
$myFunction($val);
// code
}

Answer 8

that is not the right way to do it...
first declare the function outside the loop, then call the function in the loop
function myfunction($value) {
//Do something
}
$new = array(1,2,3,4);
for($i=0;$i<=3;$i++){
$val = $new[$i];
myfunction( $val); //call function where u wanted... here (in your case)
echo $val;
}

Answer 9

You are not suposed to declare the function inside a loop...