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Home/ Questions/Q 7556967
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Editorial Team
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Editorial Team
Asked: May 30, 20262026-05-30T12:00:12+00:00 2026-05-30T12:00:12+00:00

If i’ve database my_table (id,word) as following and i’ve some posted text called $name

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If i’ve database my_table (id,word) as following

my_table (id,word)

and i’ve some posted text called $name then i want to know if $name have any word like words i’ve stored in my database my_table (id,word)

so i’m using this code

$name = "Hello lovely world"; // no bad words

$sql    = "SELECT * FROM my_table";
$result = mysql_query($sql);

$commentArray = explode(" ", $name);
$counter      = count($commentArray);
$check        = 0;

while ($row = mysql_fetch_assoc($result)) {

    for ($i == 0; $i < $counter; $i++) {

        if (strcasecmp($commentArray[$i], $row['word'])) {
            $check = 1;
        }

    }

}

if ($check == 1) {
    echo "banned";
    exit;
}
else {
    echo "passed";
}

however it always results in echo "banned"; even if i $name is clear of any bad words i’ve stored in my_table so there might be something wrong here !

anyhelp

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  1. Editorial Team
    Editorial Team
    2026-05-30T12:00:14+00:00Added an answer on May 30, 2026 at 12:00 pm

    strcasecmp returns 0 when it matches (PHP manual), so you should edit your code:

    if (strcasecmp($commentArray[$i], $row['word']) == 0)
    

    Furthermore, you have a syntax error in your for loop. It should be for ($i = 0; ...

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