I have the following code which I need to link the $asset output to another page;
Code: Select all
while($row = mysql_fetch_array($sql)){
// Build your formatted results here.
$asset=$row['asset'];
$title=$row['title'];
$type=$row["type"];
$IP=$row["IP"];
echo '<tr><td ><font color="#000000" size="2" face="Arial, Helvetica, sans-serif">'.$asset.'</font></td>Here is code taken from another page which works, however this does not work with the above as it is slightly different;
Code: Select all
echo "<TABLE BORDER=\"0\" table width=\"100%\">\n";
echo "<TR bgcolor=\"#cccccc\"><TD font colour=\"red\"><b>Asset</b></TD><TD><b>Title</b></TD><TD><b>Customer</b></TD><TD><b>Type</b></TD><TD><b>IP address</b></TD><TD><b>Reconciled</b></TD><TD><b>Edit</b></TD></TR>\n";
for($i = 0; $i < $numofrows; $i++) {
$row = mysql_fetch_array($result); //get a row from our result set
if($i % 2) {
echo "<TR bgcolor=\"#f5f7fb\">\n";
}
else {
echo "<TR bgcolor=\"#e6e6fb\">\n";
}
echo "<TD><font color=\"red\"><b><a href='viewall.php?varl=".$row['asset']."'>".$row['asset']."</a></b></font></TD><TD><font size=\"2\">".$row['title']."</TD>Please help.