I am tring to perform the following sql query to check if the name that was inserted to the form already created as taxonomy name. But I keep getting this error message:

PDOException: SQLSTATE[42S22]: Column not found: 1054 Unknown column 'test' in 'where clause': SELECT COUNT(*) FROM {taxonomy_term_data} WHERE name = test; Array ( ) in get_google_map_api() (line 79 of /Applications/MAMP/htdocs/Drupal/sites/all/modules/toodate/toodate.module).

my function is:

function get_google_map_api($item){
    $city_name = $item;
    $chack_if_is_taxonomy = (db_query("SELECT COUNT(*) FROM {taxonomy_term_data} WHERE name = ".$city_name));       
  • you error does not seem to match the code..? test column is not there in the where clause which u might have entered earlier.
    – harshal
    Commented Sep 13, 2013 at 4:27
  • You are right, I've edited my text
    – Dori
    Commented Sep 13, 2013 at 4:39
  • is the error still occuring ? Pls update the error which correlates to the code.
    – harshal
    Commented Sep 13, 2013 at 4:40
  • Yes the error is still occurring. I did update the error
    – Dori
    Commented Sep 13, 2013 at 4:43
  • What is test ? is it a column or a variable ?
    – harshal
    Commented Sep 13, 2013 at 4:47

2 Answers 2


Quick fix would be adding the missing single quotes around your variable. Forgetting the quotes can throw odd errors from this function, such as mixing a string for a column, like in your case.

There are less error prone ways to make short queries like this. See also Molot's answer.

  • Thanks, But it didn't show what I wanted to get... I'll post another question about it.
    – Dori
    Commented Sep 13, 2013 at 5:02

Don't use db_query() unless you really have to. Build your queries with db_select(). It's harder to make SQL errors using it. Your query would look like this:

// SELECT COUNT(*) FROM {taxonomy_term_data} WHERE name = $city_name
$chack_if_is_taxonomy = db_query('taxonomy_term_data', 'ttd');
$chack_if_is_taxonomy->addField('ttd', 'COUNT(*)', 'count');
$chack_if_is_taxonomy->condition('name', $city_name, '=');

As you can see, no place to make SQL errors, and PHP bugs should be easier to spot in your favourite IDE.

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