Allow where clause LIKE variable subtitution for '%?%'

[tamsler]

Allow where clause LIKE variable substitution for '%?%'

e.g. db.query().select("contact").from("contacts_t").where("contact LIKE '%?%'", ['foo']).execute(function(error, rows) { ...

Right now, this results in an:

"err":{"message":"Wrong number of values to escape"

[tomasikp]

db.query().select("contact").from("contacts_t").where("contact LIKE ''+db.escape('%'+'foo'+'%')+''').execute(function(error, rows) { ...