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I have a simple problem I need to solve, and list functions are my current attempt to do so. I have a view that generates almost what I need, but in certain cases there are duplicate entries that make it through when I send in edge-case parameters.

Therefore, I am looking to filter these extra results out. I have found examples of filtering, which I am using (see this SO post). However, rather than generate HTML or XML or what-have-you, I just want a regular ol' view result. That is, the same kind of object that I would get if I queried CouchDB without a list function. It should have JSON data as normal and be the same in every way, except that it is missing duplicate results.

Any help on this would be appreciated! I have tried to send() data in quite a few different ways, but I usually get that "No JSON object could be decoded", or that indices need to be integers and not strings. I even tried to use the list to store every row until the end and send the entire list object back at once.

Example code (this is using an example from this page to send data:

function(head, req) { 
    var row; var dupes = [];
    while(row=getRow()) { 
        if (dupes.indexOf(row.key) == -1) { 
            dupes.push(row.key); 
            send(row.value); 
        } 
    }; 
}

Lastly, I'm using Flask with Flask-CouchDB, and I'm seeing the aforementioned errors in the flask development server that I'm running.

Thanks! I can try to supply more details if need be.

Community
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Paragon
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1 Answers1

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Don't you need to prepend a [, send a , after each row value except the last, and end with ]? To actually mimic a view result, you'd actually need to wrap that in a JSON structure:

{"total_rows":0,"offset":0,"rows":[<your stuff here>]}
djc
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