Compare tables and fetch data

suguna

Hi,

I am new to php.I have a php file which compares tables and fetches data from the database.But I am getting an error.(Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in C:\wamp\www\rep28old\camdisp1.php on line 68)I have given the coding below.Pl help me to proceed further.

Thanks in advance.
<?php
session_start(); // start up your PHP session!

?>

<body>
<?php
mysql_connect("localhost","root","") or die('cannot connect'.mysql_error());
mysql_select_db("dialer");
?>
<table>
<form method=post action='<?php $SERVER[PHP_SELF] ?>' name=condisp>
From Date--------
To Date---------
Campaign---------
<?
$res = mysql_query("show tables");
$tab=$GET['tab'];
echo "<select name=tablename value= size=15>";
echo "<option value='All' Selected>All</option>";
while($res1= mysql_fetch_row($res))
{
if($res1[0]==@$tab){
echo "<option selected value='$tab'>$tab</option>"."<BR>";
}else{
echo "<option value='$res1[0]'>$res1[0]</option>";}
}
echo "</select>";
?>
</td></tr>
<tr><td> </td><td> </td></tr>
<tr><td>   <input type=submit name=submit value=Report></td></tr>
</form>
</table>
<?php
$tdate2 = $POST{'tdate'};
$fdate2 = $POST{'fdate'};

$submit = $_POST{'submit'};

$SESSION['fdate']=$fdate;
$SESSION['tdate']=$tdate;
$tname = trim($POST{tablename});
$fdate = trim($POST{fdate});
$tdate = trim($_POST{tdate});
if($submit == 'Report')
{
if($tname == 'All'){
$res =mysql_query("SELECT count( * ) , a.disposition, b.disp_name FROM $tname a, reports.disposition b WHERE a.disposition = b.disp_code and date(lastupdated) >='$fdate' and date(lastupdated) <= '$tdate' GROUP BY a.disposition");

}else{	
	$res = mysql_query("SELECT count( * ) , a.disposition, b.disp_name FROM $tname a, reports.disposition b WHERE a.disposition = b.disp_code and date(lastupdated) >='$fdate' and date(lastupdated) <= '$tdate' and tablename = '$tname' GROUP BY a.disposition");}

$r= mysql_query("select agent from $tname");
$num = mysql_num_rows($r);
$num1 = mysql_num_rows($res);
echo "<table width=100% cellspacing=0 cellpadding =0 border=1>
<tr><td>Disposition Report for</td><td> $tname [Records : $num]";
if($ddate <> 'ALL'){
<tr class=h4><td>Disposition Name</td><td align=center>Disposition ID</td><td align=center>Count</td></tr>";
if ($num1 =="0")
{
echo "<tr align=left><td>No Records Dialed</td><td align=center></td><td align=center></td></tr>";

}else{
while(list($count,$d_id,$d_name,$a_name) = mysql_fetch_row($res))
{
echo "<tr align=left><td>$d_name</td><td align=center>$d_id</td><td align=center>$count</td></tr>";
}
}
echo "</table>";
}
else{
}
echo "</table>";
?>
</form></table>

</div>
</body>
</html>

laserlight

To debug a "supplied argument is not a valid MySQL result resource" error:
1. Check your database server connection.
2. Check your database selection.
3. Check the relevant SQL statement.

NogDog

Nevermind, my eyes were playing tricks on me.

krotkruton

First, read the guide laserlight's quote: How To Ask Questions The Smart Way
Next, read some of the forum faq's about how to post... the big thing is if you are posting code, you need to surround it with open and close php tags. These are [ + PHP + ] and [ + \ + PHP + ] (without the plus signs and spaces), or select all the code in your post and click the button that looks like a piece of paper with php written on it. It will make your code look like this:

<?php //Here's my code ?>

Also, don't take that advice as meaning you need to re-post all that code, just edit your original post and put those tags in so it's easier to read. You'll probably get a better response that way.

suguna

Hi,

Sorry! Now i have reframed the question.

There are 2 databases db1 and db2 in my program.db1 has a disposition table which has disposition name and disp id. db2 has some tables which have agent id,date,disposition etc as fields.Now when the user selects a table and a period of time it should compare the tables in db1 and db2 and show the disposition name and the total no of calls in the selected table.But i am not getting the result.Pl help me to fix the problem.

if($submit == 'Report')
{
$q = mysql_query("show tables") or die ("could not run Query");

while($q1 = mysql_fetch_row($q)) 
{

if($table =='All'){
$q2 = mysql_query("SELECT count( * ) , a.disposition, b.disp_name FROM $q1[0] a, reports.disposition b WHERE a.disposition = b.disp_code and date(lastupdated) >='$fdate' and date(lastupdated) <= '$tdate' GROUP BY a.disposition");

}else{
$q2 = mysql_query("SELECT count( * ) , a.disposition, b.disp_name FROM $q1[0] a, reports.disposition b WHERE a.disposition = b.disp_code and date(lastupdated) >='$fdate' and date(lastupdated) <= '$tdate' GROUP BY a.disposition");
}
while($q3 = mysql_fetch_row($q2))
{
echo "";}
}

Thanks in advance

laserlight

There are 2 databases db1 and db2 in my program.db1 has a disposition table which has disposition name and disp id. db2 has some tables which have agent id,date,disposition etc as fields.Now when the user selects a table and a period of time it should compare the tables in db1 and db2 and show the disposition name and the total no of calls in the selected table.But i am not getting the result.Pl help me to fix the problem.

Um, why do you need two databases?