Displays username, but not maiden or last name.

[Scotty13]

This just displays the username aka first name. I want it to also display the member’s last name and maiden name if they have one. /////// Mechanism to Display Real Name Next to Username - real name(username) //////////////////////////if ($firstname != "") {$mainNameLine = "$firstname $maidenname $lastname ($username)";$username = $firstname;} else {$mainNameLine = $username; $maidenname; $lastname;} I want to display maiden and last name along with the user name. I've tried everything I could think of. Any suggestions? Thanks,Scott

[ckrudelux]

To start with you can't use end of line like this.

$mainNameLine = $username; $maidenname; $lastname;

Should be like this.

$mainNameLine = $username . $maidenname . $lastname;

Also you maybe want some space between them?

$mainNameLine = $username ." ". $maidenname ." ". $lastname;

This is how you could add everything to one variable:

//If firstname is not null set firstname else set usernameif($firstname)$mainNameLine = $firstname;else$mainNameLine = $username; //If maidenname is not null add maidennameif($maidenname)$mainNameLine .= " ".$maidenname; //If lastname is not null add lastnameif($lastname)$mainNameLine .= " ".$lastname; //If firstname is not null add usernameif($firstname)$mainNameLine .= " (".$username.")";

" .= " adds to the current value while " = " set the value.

Edited May 6, 2012 by ckrudelux

[birbal]

$mainNameLine = $username ." ". $maidenname ." ". $lastname;

it is fine. but it is best to not to use concatenate oprator where possible. it makes the code harder to read

$mainNameLine = "$username $maidenname $lastname";

you can also interpolate variables for better readbility

$mainNameLine = "{$username} {$maidenname} {$lastname}";

the obvious usage of concatenate oprator is when you need directly put the return value of a function in string or need to concat constant in string also if there is no sureity of existance of a variable it best to use isset(). without isset() if it does not exist it will throw a notice and will try to convert null to bool which behvaour is unpredictable. so it is avoidable

Edited May 6, 2012 by birbal

[ckrudelux]

I guess this is a matter of taste but I would say that this makes the code harder to read. Using the {} makes it better but still messy.

$mainNameLine = "$username $maidenname $lastname";

Why? Well then you start mixing other content in this string the variables gets harder to find by using the concatenate you make a clear mark on there data is located. Example:

"Hello $username nice to see you again, you have $numMessages messages in you inbox latest one from $from do you want to view it?"

"Hello ". $username ." nice to see you again, you have ". $numMessages ." messages in you inbox latest one from ". $from ." do you want to view it?"

In the case of the this particular operation you are right in that your use case is more readable but in longer more complected once I prefer the one I used like the one in the example.

[Scotty13]

Tried everything above and no errors, still just the username.

[birbal]

that is where interpolation make it easier, decent text editor distinguish the variable among the string by formatting it differently

[birbal]

Tried everything above and no errors, still just the username.

can you post your current code?

[Scotty13]

It maybe the myMembers_table.php I get this error when I test it... Success in database CONNECTION..... no TABLE created. You have problems in the system already, backtrack and debug! <?php require_once "connect_to_mysql.php"; print "Success in database CONNECTION.....<br />"; $result = "CREATE TABLE myMembers ( id int(11) NOT NULL auto_increment, username varchar(255) NOT NULL, firstname varchar(255) NOT NULL lastname varchar(255) NOT NULL, maidenname varchar(255) NULL, ) "; if (mysql_query($result)){ print "Success in TABLE creation!...... <br /><br /><b>That completes the table setup, now delete the file <br /> named 'create_Table.php' and you are ready to move on. Let us go!</b>"; } else { print "no TABLE created. You have problems in the system already, backtrack and debug!"; } exit(); ?> Thanks, Scott

[Ingolme]

You can use mysql_error() to see what the problem is. I think the problem is that you have a comma at the end of this line: maidenname varchar(255) NULL,