coderphp

No I have checked it Barand it gives no error,instead the data is shown but not in table but in junk form...

here is the query Kimi2k if($country=mysql_real_escape_string($_POST['country']) or DIE('please choose a country')) { $query=mysql_query("SELECT * FROM `info` WHERE `country` LIKE '$country'"); } it returns no error Barand

The problem is the data is shown but not in table only one column is shown in table and the rest are shown in junk... I don't understand why is this so??

Howdy friends, I am trying to display multiple rows from MySQL in Html table and I am using while() loop for that purpose but its not working can any one help me... here's the script.. echo " <h1><strong><center><font color='#663300'>Search Result/s for Name</font></center></strong></h1>"; echo "<table border='1' align='center' font color='#663300'> <tr> <th>Name</th> <th>City</th> <th>country</th> <th>Zimmedar/Sub Ordinate</th> <th>Course</th> </tr>"; while($far = mysql_fetch_array($query)) { echo "<tr class=\"gray\" font color='#663300'>"; echo " <td align='center' width='15%' font color='#663300'>".$far['name']."</td> "; echo " <td align='center' width='20%' font color='#663300'>".$far['city']."</td> "; echo " <td align='center' width='25%' font color='#663300'>".$far['country']."</td> "; echo " <td align='center' width='20%' font color='#663300'>".$far['z/s']."</td> "; echo " <td align='center' width='20%' font color='#663300'>".$far['course']."</td> "; echo "</tr>"; echo"</table>";} but the data is not displayed in table:-((