Question :
$idofdb = "25";
$myArray = array();
foreach ( $fthi as $codigo ) {
$myArray[] = $codigo['ImoCodigo'];
}
$checkValues = array_values($myArray);
$checkString = implode(',', $checkValues); // 3,4,5,6,7,8,9,10,11,12,13,18,19,14,15,2
I’d like to know how to compare $idofdb
with $checkString
to see if $idofdb
is registered in the database. As you can see, the output of $checkString
is a string of numbers separated by commas.
Answer :
You do not need all this, just use the in_array()
function directly in array :
in_array("25", $fthi)
Or if you feel you should do something else for some reason:
in_array("25", $checkValues)
See running on ideone .
Instead of you transforming an array into a string with comma-separated numbers you can use the in_array :
<?php
$os = array("Mac", "NT", "Irix", "Linux");
if (in_array("Irix", $os)) {
echo "Tem Irix";
}
if (in_array("mac", $os)) {
echo "Tem mac";
}
?>
Look at the PhpFiddle showing execution.
As friends have said, this is not necessary. But if you still want to check in string , you can use strpos
$pos = strpos($checkString, $idofdb);
if($pos == false) {
echo 'Não tem cadastrado';
}else{
echo 'A string foi encontrada na db';
}