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Allow where clause LIKE variable subtitution for '%?%' #69

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tamsler opened this Issue May 6, 2012 · 1 comment

2 participants

@tamsler
tamsler commented May 6, 2012

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
tomasikp commented May 8, 2012

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

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