PHP calling out index positions in an array [duplicate] - php

Question

This question already has answers here:
How to output which indexes ive found an element in an array
(4 answers)
Closed 2 years ago.
I've got a bit of a problem, I have 2 inputs, one to type words, the other to find a keyword in the word array,
My code looks like this
<form class="" action="sida3.php" method="post" >
<label for="ord">Ord</label>
<input type="text" name="ord" value="">
<label for="sök">Sökord</label>
<input type="text" name="sök" value="">
<button type="submit" name="button">Submit</button>
</form>
if (isset($_POST['ord'])){
$name = $_POST['ord'];
$searchWord = $_POST['sök'];
$nameArray = (explode(" ", $name));
foreach ($nameArray as $key => $value) {
if ($value === $searchWord) {
echo "The word ". $value .' was found ' . substr_count($name, $value) . " times";
echo "<br>";
echo "The word ". $value .' is found in position: ' . array_keys($nameArray, $value);
break;
}
}
}
The problem I have is with this line echo array_keys($value, $nameArrray);
Using array_search($value,$nameArray) results to the following output
The Word Hey is found in position: 1
I can get it to call out the position of the index once before it stops, I want it to call out all the index positions, In this cause it would be 1 and 3
It might also have to do with the break; but then the problem remains that the break prevents the whole string from repeating itself, which is why I added it
This is what the output looks like (If you type the words "hi Hey hi Hey" in the "ord" input)
Array ( [0] => hi [1] => Hey [2] => hi [3] => Hey )
The word Hey was found 2 times
Warning: array_keys() expects parameter 1 to be array, string given in C:\wamp64\www\labb-1a-php-sidor\sida3.php on line 39
The desired output should look like this
Array ( [0] => hi [1] => Hey [2] => hi [3] => Hey )
The word Hey Was found 2 times
The Word Hey is found in position: 1 3

Answer 1

In your original version you had the parameters the wrong way round in your input to array_keys - the array must be first, followed by the search value. (The version you're showing after editing the question wouldn't produce the error you mentioned, but the first one would.)
Also, array_keys returns an array and you can't echo that directly (PHP has no idea how you actually want it to appear visually), so you'll need to implode it.
foreach ($nameArray as $key => $value) {
if ($value === $searchWord) {
echo "The word ". $value .' was found ' . substr_count($name, $value) . " times";
echo "<br>";
echo "The word ". $value .' is found in position(s): ' . implode(",", array_keys($nameArray, $value));
break;
}
}
Working demo: http://sandbox.onlinephpfunctions.com/code/a1980664a7b4f8e27f7c2a1aa04ebabee5379037
P.S. You can make this a lot more efficient, there's no need for looping or substrings. Simply count the number of matching keys. You can replace all the code I showed above with just this:
$positions = array_keys($nameArray, $searchWord);
echo "The word '". $searchWord . "' is found ". count($positions) . " time(s), in position(s): " . implode(",", $positions);
Demo: http://sandbox.onlinephpfunctions.com/code/7251b2f87b7355acf5c9242cb6ceba2ea08a9b98

Answer 2

Your code at
echo '...' . array_keys($nameArray, $value);
will not be able to print because array_keys will return an array of keys (or indices) instead of a value (integer or string).
To print out an array, try one of the following approaches.
var_dump(array_keys(...));
print_r(array_keys(...));
foreach(array_keys(...) as $value) {
echo $value;
}