WhereClause returns all objects when matching on unique string

simon-archer · March 31, 2016, 7:38am

Hi there,
Ive have a ‘Campaign’ table of objects with a column called ‘referralCode’. This referralCode is a uniquely generated string, such as - ‘7541a545-6780-49bb-a18a-82452ca7f7fd’
In my iOS app, I am trying to search for and find the Campaign object, matching this referralCode value.
My code is like…


let query = BackendlessDataQuery()
 let whereClause = "referralCode = '7541a545-6780-49bb-a18a-82452ca7f7fd'"
 query.whereClause = whereClause
 
 backendless.data.of(Campaign.ofClass()).find(
 { (result: AnyObject!) -> Void in
 print(result.currentPage().count)
 },
 error: { (fault: Fault!) -> Void in
 print("Server reported an error: \(fault)")
 })

I know for a fact that there is only 1 object in my table that matches that query, but the ‘currentPage()’ results returns all objects (the count is 10)

Unsure why that could be… any ideas?
Thanks!

P.S. AppID = ‘29100CD0-A42D-0188-FFEC-947AB5CA5900’ , if that helps anything

vyacheslav-vdovichenko · March 31, 2016, 9:47am

Hi Simon,
Please, try:

let whereClause = "referralCode = \'7541a545-6780-49bb-a18a-82452ca7f7fd\'"

simon-archer · March 31, 2016, 9:51am

no luck… still getting all the objects back from the table

vyacheslav-vdovichenko · March 31, 2016, 10:10am

Please send also secretKey of your app to support@backendless.com - we investigate it and find a solution asap

simon-archer · March 31, 2016, 10:14am

Thanks, which secret key are you wanting specifically? for iOS?

vyacheslav-vdovichenko · March 31, 2016, 10:19am

yes - for iOS

simon-archer · March 31, 2016, 10:20am

Mail sent. Thanks!

vyacheslav-vdovichenko · March 31, 2016, 10:41am

Ok, you created ‘query’, but didn’t use it.

Please, change your line 594:

backendless.data.of(Campaign.ofClass()).find(query,

response: …

vyacheslav-vdovichenko · March 31, 2016, 10:54am

Here is my sample:

    func findCampaignByReferralCode() {
        
        let dataQuery = BackendlessDataQuery()
        dataQuery.whereClause = "referralCode = \'7541a545-6780-49bb-a18a-82452ca7f7fd\'"
        
        var error: Fault?
        let bc = backendless.data.of(Campaign.ofClass()).find(dataQuery, fault: &error)
        if error == nil {
            let campaign = bc.data[0] as! Campaign
            print("Campaign have been found: \(bc.data) -> &lt;\(campaign.referralCode)&gt;")
        }
        else {
            print("Server reported an error: \(error)")
        }
    }

and a log:
http://support.backendless.com/public/attachments/0f653bfdebdefca655bbe51002c19f3b.png</img>

simon-archer · March 31, 2016, 11:03am

Oh my… Cant believe I got tripped on such a rookie mistake!!!

Thanks for helping pointing that out. Was indeed missing the ‘query’ in the find call.

Thanks!!