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Home/ Questions/Q 6129939
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Editorial Team
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Editorial Team
Asked: May 23, 20262026-05-23T16:48:02+00:00 2026-05-23T16:48:02+00:00

I’m a new comer to Linux/Ubuntu so if I make any silly mistake please

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I’m a new comer to Linux/Ubuntu so if I make any silly mistake please forgive me.
My code works fine in XP but not in Ubuntu 11.04 (Same PHP/MySQL version)

<?php
  if (!isset($_SESSION['uid'])) {
    header('Location:index.php');
    exit;
  } else {
    if (file_exists("profile_pic/".$_SESSION['uid'].".jpg")) {
      echo '<a href="'.$SitePath.'profile.php?id='.$_SESSION['uid'].'"><img src="'.$SitePath.'profile_pic/'.$_SESSION['uid'].'.jpg" alt="'.$_SESSION['name'].'" title="Me" border="0"/></a>';
    } else {
  echo '<a href="'.$SitePath.'profile.php?id='.$_SESSION['uid'].'"><img src="'.$SitePath.'profile_pic/nopic.jpg" alt="'.$_SESSION['name'].'" title="Me" border="0" /></a>';
  }  
?>

If a file exists with name of $_SESSION['uid'].".jpg" then show it, else show an default image.

The error I’m getting

Catchable fatal error: Object of class mysqli_stmt could not be converted to string

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  1. Editorial Team
    Editorial Team
    2026-05-23T16:48:03+00:00Added an answer on May 23, 2026 at 4:48 pm

    You’re not showing any database operations, but most likely you’ve got something like:

    $_SESSION['uid'] = mysqli_query("SELECT uid FROM ...");
    

    wherever you’re initializing your session data. This is incorrect. The query functions return a statement handle, from which you can fetch the data. In pseudo-code, it’d be:

    $result = mysqli_query(...);
    $row = $result->fetch();
    $_SESSION['uid'] = $row['uid'];
    
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