Jump to content

Recommended Posts

Hello. How can I add a column without effects the query?

 

For example, when I add another column in SELECT, it produces each output as double.

I want to add another column because I want to create an if statement with it in the while output.

 

SELECT Users.id, Posts.post, Users.username, Posts.post_id, Posts.karma_up, Posts.karma_down

FROM Posts, Users, Friends

Link to comment
https://forums.phpfreaks.com/topic/241512-select/
Share on other sites

$friendfeed2 = 
"SELECT dbUsers.id, dbPosts.post, dbUsers.username, dbPosts.post_id, dbPosts.karma_up, dbPosts.karma_down 
FROM dbPosts, dbUsers, dbFriends, dbKarma
WHERE (dbPosts.post_id NOT IN (SELECT post_id FROM dbDownvote WHERE user_id = $user)
AND dbPosts.username_id = dbUsers.id
AND dbFriends.user_id = $user 
AND dbFriends.friend_id = dbPosts.username_id
AND dbFriends.status = 1) 

OR 

(dbPosts.post_id NOT IN (SELECT post_id FROM dbDownvote WHERE user_id = $user)
AND dbPosts.username_id = dbUsers.id
AND dbFriends.friend_id = $user 
AND dbFriends.user_id = dbPosts.username_id
AND dbFriends.status = 1) 

UNION SELECT dbUsers.id, dbPosts.post, dbUsers.username, dbPosts.post_id, dbPosts.karma_up, dbPosts.karma_down 
FROM dbPosts, dbUsers, dbPosts_share, dbFriends
WHERE (dbPosts.post_id NOT IN (SELECT post_id FROM dbDownvote WHERE user_id = $user)
AND dbPosts.username_id = dbUsers.id 
AND dbPosts_share.user_id = dbFriends.friend_id 
AND dbFriends.user_id = $user 
AND dbPosts_share.post_id = dbPosts.post_id
AND dbFriends.status = 1)

OR

(dbPosts.post_id NOT IN (SELECT post_id FROM dbDownvote WHERE user_id = $user)
AND dbPosts.username_id = dbUsers.id 
AND dbPosts_share.user_id = dbFriends.user_id 
AND dbFriends.friend_id = $user 
AND dbPosts_share.post_id = dbPosts.post_id
AND dbFriends.status = 1)

UNION SELECT dbUsers.id, dbPosts.post, dbUsers.username, dbPosts.post_id, dbPosts.karma_up, dbPosts.karma_down 
FROM dbPosts, dbUsers
WHERE dbPosts.username_id = dbUsers.id 
AND username_id = $user";

 

while( $i < $row = mysql_fetch_array($friendfeed_query2))
  {
  if ($row['id'] == $user)
  {
  echo $div_mypostbox;
  }
  else
  {
  echo $div_postbox;
  }
  $karma_up = $row['karma_up'];
  $karma_down = $row ['karma_down'];
  $karma_total = $karma_up - $karma_down;
  echo "<a href=vote.php?up=" . $row['post_id'] . "><</a>";
  echo $karma_total;
  echo "<a href=vote.php?down=" . $row['post_id'] . ">></a>";
  echo $break;
  
  $post = $row['post'];
  $post_wrap = wordwrap($post, 50, "\n", true);
  echo "$post_wrap";
  echo $div_end;
  echo $div_postfooter;
  echo "Submitted by " . $row['username'] . "<br>";

  
  if ($row['id'] == $user) {
  echo "<a href=delete.php?id=" . $row['post_id'] . ">delete</a><br>";
  }
  echo  $div_end, $div_end, $div_end, $break;
  }

 

I want to add a column in, to add an if command within the loop. I want to add something similar to what I've done with " if ($row['id'] == $user)" in the 2nd quote.

 

 

Link to comment
https://forums.phpfreaks.com/topic/241512-select/#findComment-1240628
Share on other sites

In simpler terms it always gives me multiple results of each row when I have a table that is not in SELECT.

 

for e.g

 

"SELECT dbPosts.post, dbPosts.username_id

FROM dbPosts, dbUsers

WHERE dbPosts.username_id = 10";

 

The dbUsers table isnt in the SELECT section, so it gives me multiple results of each row. How do I stop this from happening.

Link to comment
https://forums.phpfreaks.com/topic/241512-select/#findComment-1240642
Share on other sites

This thread is more than a year old. Please don't revive it unless you have something important to add.

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Restore formatting

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.