[SOLVED] Need a little help

derekbelcher · March 13, 2009

I have a php login in place on my web application. On the page you go to when you login successfully, I want to display the first name (stored in the database) that is associated with that user that is logged in. Below is the code on the page. At line 20 I have attempted something but it didn't work:

<?

session_start();

if(!session_is_registered(myusername)){

header("location:officerLogin.php");

}

?>

<html>

<head>

<title>Officer's Area</title></head>

<body>

<tr>

<td height="19" colspan="3" valign="top">Welcome, <?php $_SESSION['myusername'] ?> </td>

</tr>

<tr>

When a man becomes a fireman his greatest act of

bravery has been accomplished. What he does after that is all in the

line of work. ~Edward F. Croker

</td>

</tr>

<tr>

<td width="479" height="130" valign="top" bordercolor="#000080" style="border-style: double; border-width: 5px; padding: 1px">Evaluations are now available. To complete the evaluation packet, <a href="http://www.surveymonkey.com/s.aspx?sm=yucw0wu1q3O1I_2b18iRUtug_3d_3d">Click Here</a> The Password is: ffeval09

This evaluations will be available until March 20, 2009. Please complete before that date. </td>

</td>

<tr>

</tr>

<tr>

News and such...

<hr>

Don't forget to complete your evaluations by March 20, 2009

<hr>

We are starting strong with the Relay for Life with several ideas in the planning process. We have set a goal of $3,000 this year and we need as much participation as possible to help us reach that goal.

<hr>

</td>

</tr>

<tr>

</tr>

</table> </td>

</tr>

<tr>

</tr>

<tr>

Training Info...</td>

</tr>

<tr>

Training is progressing. Brian has implemented a new training calendar, and we have had decent turn out each Thursday. All officers are encouraged to attend each training session as this shows leadership and committment to the firefighters.

</blockquote> </td>

</tr>

<tr>

</tr>

<tr>

<td height="42" colspan="3" align="center" valign="top"><a href="logout.php">Log Out of Officers Area </a></td>

</tr>

<tr>

</tr>

</table>

</body>

</html>

jackpf · March 13, 2009

Yeah, you're not actually echoing the session variable.

If you want to send something to the page you have to echo it...

ionik · March 13, 2009

<?= $_SESSION['myusername'] ?>
// OR
<?php echo $_SESSION['myusername'];  ?>

derekbelcher · March 13, 2009

Thanks the echo worked great. Now I have one more question. There is a first name associated with each username stored in the database. How can I display the name associated with myusername? I tried to echo the variable, but I don't think it is looking at the database, it is just pulling the username from the login page.

Thanks,

jackpf · March 13, 2009

Yeah, you'll need to do something like this-

mysql_query("SELECT * FROM table WHERE username='$username'");
$fetch = mysql_fetch_array($sql);
$name = $fetch['name_of_person'];
echo $name;

ionik · March 13, 2009

you will need to pull the information from your database.

This can be done by using whichever Database you are using if you are using MySQL Database take a look here http://php.net/mysqli

derekbelcher · March 13, 2009

I got this below when I inserted the code, did I miss something? Sorry, I am just beginning to learn this stuff.

Welcome,

Warning: mysql_query() [function.mysql-query]: Access denied for user 'nobody'@'localhost' (using password: NO) in /home/athens/public_html/Officers Area/officer_index.php on line 20

Warning: mysql_query() [function.mysql-query]: A link to the server could not be established in /home/athens/public_html/Officers Area/officer_index.php on line 20

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/athens/public_html/Officers Area/officer_index.php on line 21

jackpf · March 13, 2009

Yeah, you do actually need to connect to your mysql database first...I just assumed you knew that.

Why don't you google a php-mysql tutorial or something...?

derekbelcher · March 13, 2009

ok, i connected to the database, now I get an error related to the mysql_fetch_array($sql);

Welcome,

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/athens/public_html/Officers Area/officer_index.php on line 32

Thanks.

premiso · March 13, 2009

$sql = mysql_query("SELECT * FROM table WHERE username='$username'");
$fetch = mysql_fetch_assoc($sql);
$name = $fetch['name_of_person'];
echo $name;

derekbelcher · March 13, 2009

So close, but I am still unclear as to what I am supposed to do to fix the error related to the mysql_fetch_assoc. I tried to google it, but it just confuses me.

premiso · March 13, 2009

You did change "table" to your actual table name. The above is a rough example. If this still confuses you, go find a MySQL/PHP Tutorial as suggested earlier, as it shows you have never tried one before.

derekbelcher · March 13, 2009

Yeah, I had the table name right. I know it was just a rough example. I have to be honest, I have learned more from this forum than all the tutorials I have tried. Thanks for your help. Maybe you can help explain what parts of the example need to be changed to be specific to my site. That would help. I knew the table thing because I had used that part of the code elsewhere. The column name I am trying to pull is called firstname. does that help?

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[SOLVED] Need a little help

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