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get author ID from user-avatar plugin

Posted on 2011-09-08

1 solution

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Last Modified: 2012-08-14

when a member is logged into my membership site, and posts a comment on my blog, their photo gets displayed. I'd like to be able to return the id of the author logged in so I can display their photo on a page. When I view the source code, the path is:

http://www.domain.com/wp-content/plugins/user-avatar/user-avatar-pic.php?id=21

As you can see I'm using the user-avatar plugin. How can I dynamically pull the id and place that author id in the url above? I do have a plugin called exec-php so I can embed php into the page.

Any ideas?

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Question by:COwebmaster

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10 Comments

LVL 8

Expert Comment

by:PranjalShah

ID: 365070782011-09-08

I am not familiar with the user-avatar plugin. But if you have a table with the user photos then there should be relationship between the member and its photo. You can put the member id from the table in the URL. Say for the above URL the photo belongs to some member id 3 then you can have

http://www.domain.com/wp-content/plugins/user-avatar/user-avatar-pic.php?id=21&mid=3

Now you can fetch the member info from the member table with the corresponding mid

Author Comment

by:COwebmaster

ID: 365076742011-09-08

Yes, I understand that but just need the php code to pull back dynamically the author id. Any ideas on that?

LVL 8

Expert Comment

by:PranjalShah

ID: 365081462011-09-08

Check the answer at the bottom at this question..

http://www.experts-exchange.com/Web_Development/Web_Languages-Standards/PHP/Q_27298165.html?cid=1135#a36504888

Author Comment

by:COwebmaster

ID: 365133572011-09-09

Okay, so this is what that thread has.

However, I need to display the author image. How can I obtain that?

<a href="details.php?id={$row['id']}">read more...</a>

then on detail.php all you need is:
<?php

if( isset($_GET['id']) && !empty($_GET['id']) )
{
  mysql_connect('localhost','username','password') or die( mysql_error() );
  mysql_select_db('databaseName') or die( mysql_error() );
  $sql='SELECT * FROM TableName WHERE id='.mysql_real_escape_string($_GET['id']);
  $result=mysql_query($sql) or die('Unable to execute <br />'. htmlentities($sql,ENT_QUOTES). '<br />'. mysql_error() );

  while($row=mysql_fetch_assoc($result) ){
    //assuming you have a description field
    echo $row['description'];
  }
}
exit;
?>

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LVL 8

Accepted Solution

by:PranjalShah

PranjalShah earned 500 total points

ID: 365134122011-09-09

The $_GET variable in this example is id and in the example i showed for URL your GET variable will be mid. So get the info from the member table where memberid=$_GET['mid'] and change the query accordingly

Author Comment

by:COwebmaster

ID: 365310652011-09-13

I've requested that this question be closed as follows:

Accepted answer: 0 points for COwebmaster's comment http:/Q_27298834.html#36513357

for the following reason:

thanks PranjalShah

Author Comment

by:COwebmaster

ID: 365298702011-09-13

I clicked on the wrong answer. Please remove this selection.

LVL 8

Expert Comment

by:PranjalShah

ID: 365310662011-09-13

You can not accept your own comment as a solution. That comment also comes from the link I posted

Author Closing Comment

by:COwebmaster

ID: 365328352011-09-13

Thanks!

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