how do you perform a SQL find in (?) call?

Question

This doesn't seem to be working :

@zip = %w[07005, 07034, 07035]

CardSignup.find(:all, :conditions => ["zip_code IN (?)", @zip])

=> [ ]

However, if I do a simple find_by with the second zip code, it returns a result :

CardSignup.find_by_zip_code("07034")

=> Object<x01231 ..

What am I doing wrong ?

score 3 · Accepted Answer · answered Apr 04 '11 at 11:52

Your implementation of the array is incorrect. The way you have it at the moment the commas are part of each array object:

>> %w[07005, 07034, 07035]
=> ["07005,", "07034,", "07035"]

You should implement this with the commas removed:

>> %w{07005 07034 07035}
=> ["07005", "07034", "07035"]

or:

>> [07005, 07034, 07035]
=> [07005, 07034, 07035]

So the full implementation is:

@zip = %w{07005 07034 07035}

CardSignup.find(:all, :conditions => ["zip_code IN (?)", @zip])

=> [Object<x01231 ..

score 2 · Answer 2 · answered Apr 04 '11 at 11:59

2

Don't use comma in %w[07005, 07034, 07035]

Just use  %w[07005 07034 07035]

answered Apr 04 '11 at 11:59

Ashish

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score 0 · Answer 3 · edited May 23 '17 at 09:58

0

Try:

CardSignup.find_by_zip_code(@zip)

Edit (from comments):

How to select where ID in Array Rails ActiveRecord without exception

@zip = ["07005", "07034", "07035"]
CardSignup.find(:all, :conditions => ["zip_code IN (?)", @zip])

CardSignup.find_all_by_zip_code(["07005", "07034", "07035"])

edited May 23 '17 at 09:58

Community

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answered Apr 04 '11 at 11:40

Reuben Mallaby

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How can I alter that syntax to include an `OR` statement in it? The end result I am looking for is something in the Rails/SQL syntax rather than a Rails helper. – Trip Apr 04 '11 at 11:42
WHat version of Rails are you using? – Reuben Mallaby Apr 04 '11 at 11:44
http://stackoverflow.com/questions/1441791/how-to-select-where-id-in-array-rails-activerecord-without-exception – Reuben Mallaby Apr 04 '11 at 11:45
You mixed up two different array initialisations. – Douglas F Shearer Apr 04 '11 at 11:54

how do you perform a SQL find in (?) call?

3 Answers3