How Do I Display the COUNT

tanaka99

I want to display the logid and the total count of each logid ... how do I display the actual count number.

Here's my code ...

<?php

include("dbinfo.inc.php");

mysql_connect(localhost,$username,$password);

@mysql_select_db($database) or die( "Unable to select database");

$query = "SELECT COUNT(logid) as login_count, logid FROM usercount GROUP BY logid ASC";

$result=mysql_query($query);
$num=mysql_numrows($result);

$i=0;

echo "<table border=\"2\" align=\"center\"><tr valign=\"middle\" align=\"center\"><td bgcolor=\"#c0c0c0\">UserID</td><td bgcolor=\"#c0c0c0\">Time of Login</td></tr>";

while ($i < $num) {

$logid=mysql_result($result,$i,"logid");
$logdate=mysql_result($result,$i,"logdate");

echo "<tr align=\"left\" valign=\"top\"><td>$logid</td><td>$logdate</td><td>login_count</td></tr>";

++$i;
}

echo "</table> ";

tha_mink

The thing is since you are selecting a count and grouping the result by the logid, you cannot display the login time per record. This is due to the fact that you are grouping all the logins by user. If you want to display the LAST login time for each user you would do it like this....

<?php 

include("dbinfo.inc.php"); 

mysql_connect(localhost,$username,$password); 

@mysql_select_db($database) or die( "Unable to select database"); 

$query = "SELECT COUNT(logid) as login_count, MAX(logindate) as lastlogin, logid FROM usercount GROUP BY logid ASC"; 
echo "<table border=\"2\" align=\"center\"><tr valign=\"middle\" align=\"center\"><td>Login Id</td><td>Last Login</td><td>Login Count</td></tr>"; 

while ($result = mysql_fetch_object($query)) {

echo "<tr align=\"left\" valign=\"top\"><td>$result->logid</td><td>$result->lastlogin<td>$result->login_count</td></tr>"; 


} 

echo "</table><br><font size=\"2\" face=|\"verdana,arial\"><br><br><br>"; 

?>

tanaka99

I got this result:

Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result resource in /home/virtual/site89/fst/var/www/html/dealers/userlog.php on line 18

tsinka

Hi,

the mysql_query is missing

$res = mysql_query($query);
while ($result = mysql_fetch_object($res)) {

tanaka99

Adding the new line, the error code still hits the same line:

Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result resource in /home/virtual/site89/fst/var/www/html/dealers/userlog.php on line 19

tsinka

Hi,

that error means that you have a syntax error in the query. Try to print it:

$res = mysql_query($query) or die("Error: ".mysql_error());
while ($result = mysql_fetch_object($res)) {

What does it display ?

tanaka99

All better. Thank you very much.