This is on Drupal 7.39 with Domain Access and Ubercart

When pressing "checkout" button the site returns and error which shows in the dblog as:

PDOException: SQLSTATE[42S22]: Column not found: 1054 Unknown column 'uc_payment_method_credit_checkout' in 'where clause': SELECT value FROM {variable} WHERE name =uc_payment_method_credit_checkout LIMIT 0, 1; Array ( ) in uc_domain_variable_get() (line 291 of /. . . /uc_domain.module).

Line 291 of the uc_domain_module is:

$result = db_query_range("SELECT value FROM {variable} WHERE name =" . $variable, 0, 1)->fetchField();

The column exists in the variable table and I triple checked it for an exact format match (they all exist for the variable_set commands I placed to be used in the install file)

This module is a Drupal 7 version module but is not an official project owned or maintained by anyone though it was originally written by the Drupal 6 project author and maintainer and released by him at https://www.drupal.org/node/946574. I am working with that released file as modified by "henk" - but this section is unchanged by "henk" either

The equivalent command in the Drupal 6 version is near identical

D6 version starts . . .
$result = db_result(db_query("SELECT value FROM {variable} WHERE name = '%s'", $variable));

D6 version ends . . .

I put a space in the => " . . . WHERE name = " . $variable . . ." to match the D6 version

But the error now simply reads " . . . WHERE name = uc_payment_method_credit_checkout . . . " with the space at "name = "

It seems to work -partly as it created an order entry in uc_domain_orders when i tries to test it - somehow - even though the jump to checkout failed.

This is not not like any other "column not found" issue I have searched here.

Anyone have any words of wisdom or ideas ???


3 Answers 3


Concatenating variables into an SQL query is usually a bad idea (search for SQL injection and you'll see why). The D6 version you're starting with:

$result = db_result(db_query("SELECT value FROM {variable} WHERE name = '%s'", $variable));

uses parameters, you just need to convert that to use the newer query substitution in D7. The docs for db_query() have some good examples, essentially what you need for this particular query is:

$query = 'SELECT value FROM {variable} WHERE name = :name';
$params = array(':name' => $variable);
$result = db_query($query, $params)->fetchField();
  • that's really accurate.
    – WaQaR Ali
    Aug 28, 2015 at 11:26

If you reported exactly both the error message and the code part, then the error is pretty "normal".

The D6 version
"SELECT value FROM {variable} WHERE name = '%s'"
includes simple quotes around %s, so the resulting query in the case you cited is
SELECT value FROM {variable} WHERE name ='uc_payment_method_credit_checkout' LIMIT 0, 1

At the opposite in your D7 modified version
"SELECT value FROM {variable} WHERE name =" . $variable
you didn't keep the quotes, so the resulting query becomes
SELECT value FROM {variable} WHERE name =uc_payment_method_credit_checkout LIMIT 0, 1
and without quotes SQL interprets uc_payment_method_credit_checkout as a column-name rather than a string.

This was only to explain why the error happens. In the other hand I totally agree with Clive recommendations.


i am not the author of any of this code; but Clive gave me the clue - while the question was on hold as off topic - so I reformulated the query to

 $result = db_query_range("SELECT value FROM {variable} WHERE name = :name", 0, 1, array(':name' => $variable))->fetchField();

Of course Clive's version works equally as well

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