[SOLVED] Question with MYSQL SELECT?

[$username]

In my code I am trying to SELECT username and UsrRights

When I Echo out the SQL query I do not get anything back. Why am I not getting anything echoed back?

 

 

 

include 'dbopen.php';
include 'dbconnect.php';
$username = $_COOKIE["user"];
$sql = mysql_query ("SELECT UsrRights FROM 'admin' WHERE username = '$username' ");
$sql2 = mysql_query ("SELECT username FROM 'admin' WHERE username = '$username' ");
$usrrights = $sql;
$username2 = $sql2;
?>
<div id="nav-menu">
<ul>
<li><a href="show.php">Show Cases</a></li>
<li><a href="write.php">Add New Case</a></li>
<li><a href="showusers.php">Show Clients</a></li>
<li><a href="logoff.php">Log Off</a></li>

<li><?php if ($username == $username2 && $usrrights == 120)
  echo "<a href=\"logoff.php\">Log Off</a>"; ?> </li>
</ul>
</div>
<?PHP 
echo "$username Username";
echo "$username2  username2 ";
echo "$usrrights userrights";
?>

 

 

Thank you,

Brett

[rarebit]

No need for quotes around the table name, also try using or die for errors!

$sql = mysql_query ("SELECT UsrRights FROM admin WHERE username = '$username' ") or die(mysql_error());

[$username]

test UsernameResource id #7 username2 Resource id #8 userrights

 

This is what I get when I do not use '' around the table name

 

And this is what I get for the MYSQL error

 

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''admin' WHERE username = 'test'' at line 1

[BlueSkyIS]

it looks like you expect the result of mysql_query() to be data. It's not.

 

$usrrights = $sql; // This will never be 120.

 

try something like this:

 

$result1 = mysql_query ("SELECT UsrRights FROM 'admin' WHERE username = '$username' ") or die(mysql_error());

$result2 = mysql_query ("SELECT username FROM 'admin' WHERE username = '$username' ") or die(mysql_error());

list($usrrights) = mysql_fetch_row($result1);

list($username2) = mysql_fetch_row($result2);

[$username]

Here is the value in the database.

 

password username ID secvalue UsrRights

5F4DCC3B5AA765D61D8327DEB882CF99 test 1 1 120

[stervyatnik]

Not sure I can help, but here goes!

 

When you echo something, you have to set variables apart with periods and quote in the syntax, as this:

 

$myname = 'Dan';

echo "<p>My name is ".$myname." and that's all there is to it!</p>";

 

The output will be:

 

My name is Dan and that's all there is to it!

 

Your code had - if I remember correctly - an echo like this:

 

//you generated an SQL query and loaded your result into variables like $username1 and $username2

echo "$username1 $username2 etc...";

 

You should try this with your variables:

 

//you generated an SQL query and loaded your result into variables like $username1 and $username2

//for example $username1 = 'Joe' and $username2 = 'Fred'

echo $myname1." ".$myname2." etc...";

 

Your output will be:

 

Joe Fred etc...

 

Hope that helps!

 

 

Y

[BlueSkyIS]

"When you echo something, you have to set variables apart with periods and quote in the syntax, as this;"

 

No, that is wrong.

echo "<p>My name is $myname and that's all there is to it!</p>";

 

is exactly the same as

 

echo "<p>My name is ".$myname." and that's all there is to it!</p>";

 

THE PROBLEM is that we are calling mysql_query(), which returns a mysql resource id, NOT a row of data. To get the data, we still have to call mysql_fetch_row() or similar, using the returned resource id.

 

[stervyatnik]

To clarify, using your code:

<?

include 'dbopen.php';

include 'dbconnect.php';

$username = $_COOKIE["user"];

$sql = mysql_query ("SELECT UsrRights FROM 'admin' WHERE username = '$username' ");

$sql2 = mysql_query ("SELECT username FROM 'admin' WHERE username = '$username' ");

$usrrights = $sql;

$username2 = $sql2;

?>

<div id="nav-menu">

<ul>

<li><a href="show.php">Show Cases</a></li>

<li><a href="write.php">Add New Case</a></li>

<li><a href="showusers.php">Show Clients</a></li>

<li><a href="logoff.php">Log Off</a></li>

 

<li><?php if ($username == $username2 && $usrrights == 120)

  echo "<a href=\"logoff.php\">Log Off</a>"; ?> </li>

</ul>

</div>

<?PHP

echo "$username Username";

echo "$username2  username2 ";

echo "$usrrights userrights";

?>

 

Your last three lines are the problem. They should be:

echo $username." Username";

echo $username2." username2";

echo $usrrights." " userrights";

 

Hope that helps, too!

 

[BlueSkyIS]

^^^ WRONG!

 

$sql = mysql_query ("SELECT UsrRights FROM 'admin' WHERE username = '$username' ");
$sql2 = mysql_query ("SELECT username FROM 'admin' WHERE username = '$username' ");
list($usrrights) = mysql_fetch_row($sql);
list($username2) = mysql_fetch_row($sql2);

[$username]

This is how I fixed it.

 

<?php

if (isset($_COOKIE["user"]))

setcookie("user", ($_COOKIE["user"]), time()+600);


else
header("Location:cookerr.htm");
//echo "Welcome guest!<br />";


?>
<?PHP
include 'dbopen.php';
include 'dbconnect.php';
$username = $_COOKIE["user"];

$result1 = "SELECT UsrRights FROM admin WHERE username = '$username' ";
$result2 = "SELECT username FROM admin WHERE username = '$username' ";
$query1 = mysql_query($result1);
$query2 = mysql_query($result2);
list($usrrights) = mysql_fetch_row($query1);
list($username2) = mysql_fetch_row($query2);
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<link href="./css/master.css" rel="stylesheet" type="text/css" />

</script>
<html>

<div id="nav-menu">
<ul>
<li><a href="show.php">Show Cases</a></li>
<li><a href="write.php">Add New Case</a></li>
<li><a href="showusers.php">Show Clients</a></li>
<li><a href="logoff.php">Log Off</a></li>


</ul>
</div>
<?PHP 
echo "$username Username";
echo "$usrrights  username2 ";
echo "$username2 userrights";
?>