I want to search a array and use more than one word, the problem is that I want it to only return something if all words in array exist and not just one, I'm doing the following form but it is not working:
<?php
if(isset($_POST['palavra'])):
$palavra = $_POST['palavra'];
$separa = explode(' ', $palavra);
if(in_array(array("você", "bem", "?"), $separa)):
echo "Bem e você ?";
endif;
endif;
?>
Maybe you can do something like this:
if(isset($_POST['palavra'])):
$palavra = $_POST['palavra'];
$separa = explode(' ', $palavra);
$palavrasChaves = array("você", "bem", "?");
$quaisContem = array_intersect($separa, $palavrasChaves);
sort($palavrasChaves);
sort($quaisContem);
if($quaisContem === $palavrasChaves):
echo "Bem e você ?";
endif;
endif;
I used array_intersect to check which words of the second array are found in the first. Thus, it returns which are values found in the first array, which are in the second. I apply a sort in both, for the ordering to be similar. Then I compare the two with === . If true, it's because all the words you're looking for are in the list you've separated.
So you can understand a little about array_intersect , I'll give you some examples:
array_intersect(['a', 'b'], ['a']); // ['a']
array_intersect(['a'], ['a', 'b']); // ['a'] ('a' presente no segundo está no primeiro
Another alternative is the array_diff function, you can implement it like this < sup> ( credits ) :
function in_array_all($needles, $haystack) {
return !array_diff($needles, $haystack);
}
The array_diff returns a array with all values of $needles that does not appear in $haystack , to reverse the result is used Logical Denial Helper ! .
Use this:
$palavrasChaves = ["você", "bem", "?"];
if (isset($_POST['palavra']) && $palavra = $_POST['palavra']) {
$pedacos = explode(' ', $palavra);
if (in_array_all($palavrasChaves, $pedacos)) {
echo "Bem e você ?";
} else {
echo "Não estou bem.";
}
}