• Status: Solved
  • Priority: Medium
  • Security: Public
  • Views: 1378
  • Last Modified:

Catchable fatal error: Object of class variant could not be converted to string error in PHP

I am trying some basic code, but am producing the error message in the subject line above.  It is returning this error because the result is a NULL value; however, I have tried using empty and is_null and it still produces the same error.  Here is some relevant code:

<?php
require("dbQS.php");

$strPart = CheckPost('hidPart', '');
$strCountryOfOrigin = CheckPost('txtCountryOfOrigin', '');
$intPO = CheckPost('hidPO', '');
$strMfg = CheckPost('hidMfg', '');
$strTariffCode = trim(CheckPost('txtTariffCode', ''));

$ssql = "SELECT TariffCode, ID FROM InventoryMaster1 WHERE ([PART #] = '$strPart')";
$rs = $dbc->execute($ssql);

if (!$rs->EOF)
{
      $strOldTariffCode = $rs['TariffCode'];
      $intID = $rs['ID'];
}else{
      $strOldTariffCode = "";
      $intID = "";
}

$rs->close;
$rs=null;

if (is_null($strOldTariffCode))
{
      $strOldTariffCode = "Test";
}

echo $strOldTariffCode;

The error occurs on the echo $strOldTariffCode; line.

I have replaced the is_null with empty and it produced the same error.  I have also changed it to if (trim($strOldTariffCode) == NULL) and it still produces a seperate error of:

Warning: trim() expects parameter 1 to be string, null given

Any ideas on how to get past this?  Basically, I just want the result ot be a blank value if it is a NULL value.
0
dzirkelb
Asked:
dzirkelb
  • 9
  • 5
1 Solution
 
Lukasz ChmielewskiCommented:
try:
1. adding error_reporting(E_ALL); to the top of the script
2. checking if $strOldTariffCode === NULL
3. checking is_object($strOldTariffCode)
0
 
dzirkelbAuthor Commented:
options 1 and 2 produced the same error.

Option 3 returns true, so it gives the value of $strOldTariffCode as "Test" (minus the quotes).

unfortunately I will not be able to use that code as when it does have a value, then it will change it's value to the "Test" when I don't want it to.
0
 
Lukasz ChmielewskiCommented:
Try this to see what the variable look like:

if (is_object($strOldTariffCode))
{
      var_dump($strOldTariffCode);
}
0
Free Tool: Subnet Calculator

The subnet calculator helps you design networks by taking an IP address and network mask and returning information such as network, broadcast address, and host range.

One of a set of tools we're offering as a way of saying thank you for being a part of the community.

 
dzirkelbAuthor Commented:
object(variant)#3 (0) { }  is what is produced.
0
 
Lukasz ChmielewskiCommented:
http://php.net/manual/en/function.variant-get-type.php

try this below to see what is the variant type
if (is_object($strOldTariffCode))
{
      $test = variant_get_type($strOldTariffCode);
      echo"$test";
}
0
 
dzirkelbAuthor Commented:
it returns 9
0
 
Lukasz ChmielewskiCommented:
OK, then try to compare it like this:

if(variant_get_type($strOldTariffCode) == VT_NULL)
{
  echo"this is null";
}
else{
   echo"not null !";
}
0
 
dzirkelbAuthor Commented:
This returns not null !

However, it is incorrect as it should be null.

I tested this with another variable in my code from above, $intID, which after running the query returns the value of 146553.

So, in summary, this is what is happening with the new code just provided:

$strOldTariffCode is equal to NULL
if(variant_get_type($strOldTariffCode) == VT_NULL)
{
  echo"this is null";
}
else{
   echo"not null !";
}

$intID is equal to 146553
if(variant_get_type($intID) == VT_NULL)
{
  echo"this is null";
}
else{
   echo"not null !";
}

returns not null !
0
 
dzirkelbAuthor Commented:
sorry, forgot to produce a result:

$strOldTariffCode is equal to NULL
if(variant_get_type($strOldTariffCode) == VT_NULL)
{
  echo"this is null";
}
else{
   echo"not null !";
}

returns not null !
0
 
Lukasz ChmielewskiCommented:
seems like its a type of VT_DISPATCH
http://php-manual.skryptoteka.pl/com.constants.html
try - just to check

if(variant_get_type($strOldTariffCode) == VT_DISPATCH)
{
  echo"this is pointer";
}
else{
   echo"not pointer !";
}
0
 
dzirkelbAuthor Commented:
it prints this is pointer
0
 
dzirkelbAuthor Commented:
Also,  the backed end database is sql 2000 and the server is windows 2003 running iis 6.0 if that helps at all.
0
 
dzirkelbAuthor Commented:
echo $strOldTariffCode->value
if (is_null($strOldTariffCode->value))
var_dump($strOldTariffCode->value)

all produce the following error if this helps:

Fatal error: Uncaught exception 'com_exception' with message 'Source: ADODB.Field Description: Object is no longer valid.' in D:\xxx.php:26 Stack trace: #0 D:\xxx-save.php(26): unknown() #1 {main} thrown in D:\xxx-save.php on line 26
0
 
dzirkelbAuthor Commented:
Thanks for all of your help.  I figured out that doing $strOldTariffCode = $rs['TariffCode']->value; produces the results I am looking for.
0
Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.

Join & Write a Comment

Featured Post

Cloud Class® Course: Microsoft Exchange Server

The MCTS: Microsoft Exchange Server 2010 certification validates your skills in supporting the maintenance and administration of the Exchange servers in an enterprise environment. Learn everything you need to know with this course.

  • 9
  • 5
Tackle projects and never again get stuck behind a technical roadblock.
Join Now