相同的代码不同地从数据库加载数据和使用本地数据

Question

(Yes, the title should have been more understandable but honestly I didn't know how to set it.)

What I am trying to do is creating a online dictionary of Latin. When you search a word the code (I am using some index so don't worry about selectivity) will search that name row by row inside a string (the string becomes an array using explode).

I've created a table called 'test' just to practice with this code but it works in the same way of the one of the dictionary: here the table

Before trying to load data from database I've checked if it worked locally. It works locally. Here the code:

$word_searched = 'puellam';

$string_declination = ('puella,puellae,puellae,puellam,puella,puella');
$array_declination = explode(",", $string_declination);

$check = 0;

foreach ($array_declinazion as $item) {
    if ( (strcmp($item, $word_searched)) === 0 ){
        $check++;
    }
}   

if ( $check > 0) {
    echo 'yes';
}

Then I've tryed to load the data from the database (the table 'test') using the same code. Here the code:

$query =  "SELECT * FROM test";

$result = mysqli_query($conn, $query) or die('Error running query!');

$check = 0;
$string = 'puellam';

while($row = $result->fetch_assoc()) { 
    $array_declinazione = explode(",", $row['declinazione']);
    foreach ($array_declinazione as $item) {
        if ( (strcmp($item, $string)) === 0 ){
            $check++;
        }
    }
}


if ( $check > 0) {
    echo 'yes';
}

The first code works perfectly but the second one doesn't even though it's the same way. I just don't understand why.

Answer 1

I think what you need for the second one is this:

$string = 'puellam';
// If there is a word that matches $string it will return the row
$query =  "SELECT * FROM test WHERE declinazione LIKE '%$string%'"; // demonstration purposes only. THIS IS VERY UNSAFE!!!! DO NOT USE IN PRODUCTION!!!!

$result = mysqli_query($conn, $query;

if ( $result ) {
    $check = mysqli_num_rows($result);

    if ( $check > 0) {
        echo 'yes';
    }
} else {
    die('Error running query! ' . mysqli_error($conn); // Throw errors if any - It's best to log it instead of "echoing" it
}