$result = mysql_query("SELECT * hours WHERE member='$user'");

[madjack87]

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/aci/docs/admin/hours.php on line 20

 

<?php 
$user = $_SESSION['myusername'];
$result = mysql_query("SELECT * hours WHERE member='$user'");

while ($row = mysql_fetch_array($result)){
echo $row['date'] . " " . $row['time'] . "<br />";
}

?>

 

CREATE TABLE `hours` (
  `hoursID` int(5) NOT NULL auto_increment,
  `member` varchar(20) NOT NULL default '',
  `date` date NOT NULL default '0000-00-00',
  `time` time NOT NULL default '00:00:00',
  PRIMARY KEY  (`hoursID`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=7 ;

-- 
-- Dumping data for table `hours`
-- 

INSERT INTO `hours` VALUES (4, 'aci', '2010-11-30', '14:31:39');
INSERT INTO `hours` VALUES (2, 'aci', '2010-11-30', '14:31:08');
INSERT INTO `hours` VALUES (3, 'aci', '2010-11-30', '14:31:23');
INSERT INTO `hours` VALUES (5, 'aci', '2010-11-30', '14:31:40');
        

[requinix]

http://php.net/manual/en/function.mysql-error.php

[madjack87]

It has nothing to do with my database connectivity. If there is some code I can add to give a better error please tell me how to add that to what I currently have.

[madjack87]

I added die (mysql_error());

 

this is the error it gives now.

 

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 'hours WHERE member='aci'' at line 1

 

Where member = aci the aci is pulled from the session id which is correct.

 

Any help would be appreciated.

[AbraCadaver]

http://dev.mysql.com/doc/refman/5.1/en/select.html

[kenrbnsn]

I would write the query in two lines, so debugging is much easier:

<?php
$query = "SELECT * hours WHERE member='$user'";
$result = mysql_query($query) or die("Problem with the query: $query<br>" . mysql_error());
?>

 

That said, your error is that you left out the required word "from" in the query:

<?php
$query = "SELECT * from hours WHERE member='$user'";
?>

 

Ken

[Pikachu2000]

It would be a good idea to:

 

First: remove the query string from the query execution, store it in a variable, and use the variable in the execution.

Second: check whether the query executed successfully. If yes, make sure it returned more than an empty result set. If no, echo the query with any error.

Third: make sure that any script making use of $_SESSION variables has session_start() somewhere in it before any output is generated.

Fourth: Since the query is really contingent on $_SESSION['username'] having a value, check for it, and if not set, don't execute the query.

 

session_start();
if( !isset($_SESSION['username']) ) {
   // $_SESSION variable is not set.
   echo "Session variable not set.";
} else {
   $user = $_SESSION['myusername'];
   $query = "SELECT * hours WHERE member='$user'";
   if( $result = mysql_query($query) ) {
      if( mysql_num_rows($result) > 0 ) {
         while ($row = mysql_fetch_array($result)) {
            echo $row['date'] . " " . $row['time'] . "<br />";
         }
      } else {
         // Query succeeded, but result set was empty
         echo "No Results Returned.";
      }
   } else {
      echo "<br>Query: $query<br>Failed with error: " . mysql_error();
   }
}

[jdavidbakr]

Also ... you might change your query to:

 

$query = "SELECT * from hours WHERE member='$user'";

[madjack87]

I forgot the FROM in select * from hours I had select * hours.

 

thanks for the solution.

 

Also Session start was at the top of the html page so I know it was set if the session wasnt active it would have redirected to the login page.

 

Also I knew that the session was working because I echo the username to make sure.

 

Thank you all for you help and pikachu I will be saving this page so I can look at your code and study it more as it is more professional that I currently write..