查询在php中返回0个结果，但是在oracle SQL开发人员中它返回52个结果

Question

I have a problem with with my oracle query. When I run it in PHP, it returns 0 results. If I run the same Query in Oracle SQLDeveloper, it returns 52 results.

This is the query in PHP:

SELECT name , zipcity FROM  import_persons WHERE upper(:zoekop) LIKE '%:zoekwaarde%' AND status < 3 ORDER BY name
$parameters = array(':zoekop' => $zoekop, ':zoekwaarde' => $zoekwaarde);

The query I run in SQLDeveloper, with the variables filled in manually.(Copied them from an echo in the PHP).

SELECT  name , zipcity FROM import_persons WHERE upper(name) LIKE '%Q%' AND status < 3 ORDER BY name

I can't seem to find the error, before asking, I tried to bind the params but that didn't work.

Answer 1

There are two problems here:

SELECT name , zipcity FROM  import_persons
    WHERE UPPER(:zoekop) LIKE '%:zoekwaarde%' AND status < 3
    ORDER BY name

It is good you're using bound parameters, but unfortunately only values can be bound; since :zoekop is a placeholder for a column, you'll have to use traditional string concatenation (and if the column name comes from user input, use a whitelist for security).

You can bind for :zoekwaarde , but the wildcards need to be part of the string you bind, not in the query. Your SQL will therefore look a bit like this:

SELECT name , zipcity FROM  import_persons
    WHERE UPPER(zoekop_col) LIKE :zoekwaarde AND status < 3
    ORDER BY name

Notice that the query does not include quotes - they are not required, since the binding system knows the type of the parameter you are binding in this position. Wrapping it all up, you now only have one parameter to bind:

$parameters = array(':zoekwaarde' => "%{$zoekwaarde}%", );

Answer 2

i don't know PHP so this is a bit of a guess, but i wonder if the

LIKE '%:zoekwaarde%' 

is taking the variable name as a literal string and you many be better off with something like

LIKE '%' || :zoekwaarde || '%'

Answer 3

(Posted on behalf of the OP.)

Thanks to halfer's answer, I was able to fix the problem. The fix:

$query_01 = "SELECT name , zipcity FROM  import_persons WHERE upper(name) LIKE :zoekwaarde AND status < 3 ORDER BY name";
$parameters = array(':zoekwaarde' => "%{$zoekwaarde}%", );