Hi there,
Can some one tell me why the two same codes below one without table tag works and another with table tag not? Thanks,
Below code works:
Below code does not work:Code:<form name="register" id="registerForm" action="inc/peregister.php" method="post"> <label class="labelLong" for="penewuser">Please choose a user name: </label><input type="text" name="penewuser" id="penewuser" size="24" /><span class="error">The name is taken, please choose another</span><br /> <label class="labelLong" for="penewpass">Please choose a password: </label><input type="password" name="penewpass" id="penewpass" size="24" /><span class="error">password must not be blank</span><br />
Javascript:Code:<table bordre=0><tr> <td><label class="labelLong" for="penewuser">Please choose a user name: </label></td><td><input type="text" name="penewuser" id="penewuser" size="24" /></td><td><span class="error">The name is taken, please choose another</span></td></tr> <tr><td><label class="labelLong" for="penewpass">Please choose a password: </label></td><td><input type="password" name="penewpass" id="penewpass" size="24" /></td><td><span class="error">password must not be blank</span></td></tr> ....
Code:$(document).ready(function() { /* new user name check */ $('#penewuser').blur(function() { var newName = $(this).val(); $.post('inc/peRegister.php', { formName: 'register', penewuser: newName }, function(data){ var usernameCount = data; if(1 == usernameCount){ $('#penewuser').next('.error').innerHTML = "What?"; $('#penewuser').next('.error').css('display', 'inline'); } else { $('#penewuser').next('.error').css('display', 'inline'); $('#penewuser').next('.error').innerHTML = usernameCount; } }, 'html'); }); });


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