building SQL Query From another Query in php

Question

When I Try to built Query from another Query in php code I Faced some problem
can you tell me why? :(

code :

$First="SELECT ro.RoomID,
               ro.RoomName,
               ro.RoomLogo,
               jr.RoomID,
               jr.MemberID,
               ro.RoomDescription 
          FROM joinroom jr,rooms ro 
         where (ro.RoomID = jr.RoomID)
           AND jr.MemberID = '1' ";

$sql1 = mysql_query($First);

$constract .= "ro.RoomName LIKE '%$search_each%'";
$constract="SELECT * FROM $sql1 WHERE $constract ";// This statment is Make error

thanks ,Query it is Work now ! ..

But I Faced another problem ,When I Display the Result of this Query ..

code :

$run =mysql_query ($sql);

while($runrows=mysql_fetch_assoc($run))         
{
  $RoomName=$runrows["ro.RoomName"];
  $RoomDescription=$runrows["ro.RoomDescription"];

  echo "<center><b>$RoomName</b><br>$RoomDescription<br></center>";
}

Knowledge Craving · Accepted Answer · 2010-05-15 16:53:00Z

2

Regarding the last part of the error, which you are facing, you need to remove the prefix "ro." from the right hand parts of the assignment statements.

So your actual code should have been:-

$run = mysql_query ($sql);
while($runrows = mysql_fetch_assoc($run)) {
    $RoomName = $runrows["RoomName"];
    $RoomDescription = $runrows["RoomDescription"];

    echo "$RoomName $RoomDescription";
}

Try this one above, & you should be working fine; until & unless there is some error regarding your field's existence in your database table or your database connection.

answered May 15, 2010 at 16:53

Knowledge Craving

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2 Comments

Nina Over a year ago

Thank ! it's Working well now! : )

Knowledge Craving Over a year ago

Thanks to you, for testing it quick enough & setting it as the answer.

Alec · Accepted Answer · 2010-05-15 16:09:18Z

0

You don't need a subquery to add the LIKE-part:

SELECT ro.RoomID,ro.RoomName,ro.RoomLogo,jr.RoomID, jr.MemberID,ro.RoomDescription
FROM joinroom jr, rooms ro 
WHERE ro.RoomID = jr.RoomID
  AND jr.MemberID = '1'
  AND ro.RoomName LIKE '%$search_each%'

Btw, make sure to sanitize/escape the $search_each variable.

answered May 15, 2010 at 16:09

Alec

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