Check Blob field if null or ok

Posted on 2006-07-05
Medium Priority
Last Modified: 2008-02-26
Hello experts

I need your help,  When display out (echo) blob data it occurs error if there no data.
How can I check blob field is filled or not with data (actually image)?

Seems Empty() function is not works.
Any good ideas will be appreciated. tnx

My code:

      $table = 'fnsonld.emp_emp';
      $fldblob = 'PHOTO';
      $fldid = 'EMPID';
      $valid = '16';
      if (isset($_REQUEST['t'])) $table = $_REQUEST['t'];
      if (isset($_REQUEST['f'])) $fldblob = $_REQUEST['f'];
      if (isset($_REQUEST['i'])) $fldid = $_REQUEST['i'];
      if (isset($_REQUEST['v'])) $valid = $_REQUEST['v'];

      $query = 'SELECT '.$fldblob.' FROM '.$table.' WHERE '.$fldid.'='.$valid;
      $stmt = oci_parse ($conn, $query);
      oci_execute($stmt, OCI_DEFAULT);
      $arr = oci_fetch_assoc($stmt);
      $result = $arr[$fldblob]->load();
            header ("Location: ./img/noPic.gif");
            header("Content-type: image/JPEG");
            echo $result;
      oci_close($conn); // log off
Question by:jambuul
  • 2
LVL 29

Accepted Solution

TeRReF earned 500 total points
ID: 17048664
$arr = oci_fetch_assoc($stmt);
if (count($arr) && isset($arr[$fldblob])) {
     $result = $arr[$fldblob]->load();
      header("Content-type: image/JPEG");
      echo $result;
} else {
      header ("Location: ./img/noPic.gif");

Author Comment

ID: 17048689
You are really fast.
tnx a lot
LVL 29

Expert Comment

ID: 17048695
You're welcome... (must be the amount of coffee i've had already :)

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