Date Format Problem When I use Left Join

[ayesha789]

I want result as Not Available if value is null or 0000-00-00 or 1971-01-01Hi, this is a code , which I used for showing results from 3 tables.
Lease_North is my main table.
When I join with another 2 tables using Left, if there is not value exist in table 2 and table 3 then its shows date 1-January-1971. although I used if condition to control, but its showing again and again 1971 date.
please help me out.

<?php
include("config.php");

$result = mysql_query("SELECT tb.LAPeriodStart,tb.code,Year(lp.durations),Year(lp.duratione),lp.pDate,lp.rDate,lp.siteid,lc.cDate
FROM lease_north as tb
Left JOIN limsrentpayment as lp ON tb.Code =mid(lp.SiteID,7)
Left JOIN limscheaquesdispatched as lc ON tb.code = mid(lc.SiteID,7)
WHERE tb.SiteId REGEXP '$a'");


while($row = mysql_fetch_array($result))
  {
echo "<tr>";
if($row[2]!='')
{
echo "<td   >" . $row[2] . "-" . $row[3] . "</td>";
}
else
{
echo "<td>Not Available</td>";
}
echo "<td colspan >" .date("j-F",strtotime($row[0])). "</td>";
if($row[4]!='0000-00-00')
{
echo "<td>" .date("j-F-Y",strtotime($row[4])). "</td>";
}
else
{
echo "<td>Not Available</td>";
}
echo "<td>Not Available</td>";
if($row[5]!='0000-00-00' )

{
echo "<td>" .date("j-F-Y",strtotime($row[5])). "</td>";
}
else
{
echo "<td>Not Available</td>";
}
if($row[7]!='')
{
echo "<td>" .date("j-F-Y",strtotime($row[7])). "</td>";
}
else
{
echo "<td>Not Available</td>";
}
echo "<td>Not Available</td>";
if($row[7]!='')
{
echo "<td> Complete </td>";
}
else
{
echo "<td> Incomplete </td>";
}
echo "</tr>";
	 //$b =$row['SiteId']; 

  }

?>

[ayesha789]

Please reply ASAP

[ayesha789]

Please check my problem

[ayesha789]

Is it very difficult?

[ayesha789]

I use one table for storing all values and my problem solved.
But using left join its a problem.

[JRM]

I use one table for storing all values and my problem solved.
But using left join its a problem.

You are using a numeric array, which makes it very difficult to follow if one does not know the table structures. (even if you do!)
Could it be that you are referencing the the wrong cell?
I prefer using associative arrays for this reason.