echo a number in php - php

Question

I am trying to learn php, and I am playing around with while loops. I was wondering how to print out a specific number in an array in php. Fx:
$a = [1,3,5,7,9,11,13];
$s = 3;
while($a == 3) {
echo $s.' is in the row';
$a++;
}
In this example I would like to run through the $a and see if 3 exist there. If it does it has to echo '3 is in the row' I tried to make a while loop, but it is not correct. Can anyone see what I am doing wrong? Just to say it, I think it is very wrong, but I don't know how to solve it, if I have to use the while loop?
Best Regards
Mads

Answer 1

Your while condition reads: "While the value of $a equals 3", but $a is an array, so its value can't ever be 3. The loop will never be executed. In PHP, we would write:
if (in_array($s, $a))
echo $s, ' was found in the array';
Or, if you insist on writing loops:
foreach ($a as $key => $value)
{
if ($value == $s)
{
echo $s, ' was found at offset ', $key;
break;//end terminate loop
}
}
Of course, you could also write:
for ($i=0, $j=count($a);$i<$j;++$j)
{
if ($a[$i] == $s)
{//you could move this condition to the loop itself, even
echo $s, ' found in array at offset ', $i;
break;
}
}
You can, if you want use a while loop, too, but that wouldn't be the best choice for your particular case. Just read through the manual on php.net. There are many, many array_* functions available, and there are many ways to iterate over your data.
Another worry is your using the array name as a sort-of C-style pointer: $a++; in C, an pointer can be incremented to set it to point to the next value in an array (if the new memory address is valid, and the pointer is valid, and all of the other things you have to worry about in C). PHP does not work this way. An array isn't really an array: it's a hash map. incrementing an array, therefore, is pointless and most likely to be a bug. The for loop is the closest you can get to traversing an array using the ++ operator.

Answer 2

You're looking for in_array. This checks if a value exists in an array, in the form of:
in_array ( mixed $needle , array $haystack )
So, in your case, you'd want to do:
$a = [1,3,5,7,9,11,13];
$s = 3;
if (in_array($s, $a)) {
echo $s.' is in the row';
}

Answer 3

foreach($a as $b) {
if($b == 3)
echo $b.' is in the row';
}

Answer 4

Modify slightly your code changing while condition:
$a = array(1,3,5,7,9,11,13);
$s = 3;
$counter = 0;
while($counter < count($a)) {
if ( $a[$counter] == $s )
echo $s.' is in the row';
$counter++;
}
Added counter to iterate through while loop until end of array.
count() method returns number of items in array.
This solution prints all occurences of your number.
To have better code, change names of variables:
$numbers = array(1,3,5,7,9,11,13);
$target = 3;
$counter = 0;
while($counter < count($numbers)) {
if ( $numbers[$counter] == $target )
echo $target.' is in the row';
$counter++;
}

Answer 5

There are two ways to do it,
First, you can loop through all items in the array using a foreach() loop.
That way, you can go through them all and if you have multiple conditions, it makes your code a bit more readable.
And example of that loop is like this:
foreach($array as $array_item) {
if($array_item === 3) {
echo "3 is in the array";
}
}
The alternative is to use a built in function to find if something is in the array. THis is probably much faster, though I haven't benchmarked the difference.
if(in_array(3, $array)) {
echo "3 is in the array";
}

Answer 6

you can use
array_search ,in_array , and forearch or for loops to itertate through the array.

Answer 7

For learning purposes
$a = [1,3,5,7,9,11,13];
$s = 3;
for($i=0;$i<count($a);$i++)
{
if($a[$i]==$s){
echo $s.' is in the row';
}
}
of course in real life
if (in_array(3, $a)) {
// Do something
}
would be better;

Answer 8

<?php
$a = [1,3,5,7,9,11,13];
$s = 3;
for($a=0;$a < 20; $a++)
{
while($a == 3) {
echo $s.' is in the row';
//$a++;
}
}
?>