PHP Code:

    <form>
    <table border="0">
    <tr><td>Gebruikersnaam: &nbsp; &nbsp;</td><td><input type="text" maxlength="25" name="username" id="username" onkeyup="javascript:countChars1('counter_number1')" /><span id="counter_number1">25</span> <span id="username_status"> </span></td></tr>
    <tr><td>E-mail:</td><td><input type="text" maxlength="45" name="email1" id="email1" onkeyup="javascript:countChars2('counter_number2')" /><span id="counter_number2">45</span></td></tr>
    <tr><td>Herhaal e-mail:</td><td><input type="text" maxlength="45" name="email2" id="email2" onkeyup="javascript:countChars3('counter_number3')" /><span id="counter_number3">45</span> <span id="email_status"> </span></td></tr>
    <tr><td>Wachtwoord:</td><td><input type="password" maxlength="30" name="password1" id="password1" onkeyup="javascript:countChars4('counter_number4')" /><span id="counter_number4">30</span> <span id="password_status"> </span></td></tr>
    <tr><td>Herhaal wachtwoord:</td><td><input type="password" maxlength="30" name="password2" id="password2" onkeyup="javascript:countChars5('counter_number5')" /><span id="counter_number5">30</span></td></tr>
    <tr><td></td><td><input type="checkbox" value="1" /> Ik ga akkoord met de <a href="#" onclick="showhide('av', 'block'); return false" >algemene voorwaarden</a>.</td></tr>
    <tr><td></td><td><input type="button" name="Button" value="Registreren" onclick="javascript:attach_file('mysql_insert.php?username=' + username.value + '&email1=' + email1.value + '&email2=' + email2.value + '&password1=' + password1.value + '&password2=' + password2.value); show_status('Uw gegevens aan het checken..');"> <span id="status"></span></td></tr>
    </table>
    </form> 


In het bovenstaande scriptje gaat iets fout waardoor show_status niet goed gaat, iemand een idee?