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$.csv.parsers.splitLines(csv) does not work #27
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From dirk.ste...@wondergraphs.com on April 07, 2013 11:20:26 Checked the source and it's a simple problem of an empty options.state object. Line 278: if(!options.state.rowNum) When I pass in an option like this it works: |
From evanpla...@gmail.com on April 07, 2013 21:25:38 I'll take a look at it. I added more state tracking on the last update and probably just missed the default state object creation on the splitLines method. Thank you for the feedback. |
From evanpla...@gmail.com on April 22, 2013 17:59:26 If you want a quick and dirty fix. Just add a 'state' object literal with rowNum and colNum both initialized to 0. To see what I mean, just take a look at some of the other functions. The proper fix would be to make splitLines so it inherits its state from the function arguments with defaults as a fallback. |
From evanpla...@gmail.com on December 09, 2013 15:32:06 Status: Accepted |
Can you specify how this is done? I just add |
From dirk.ste...@wondergraphs.com on April 07, 2013 11:09:09
Returns:
Uncaught TypeError: Cannot read property 'separator' of undefined jquery.csv-0.71.min.js:27
$.csv.parsers.splitLines jquery.csv-0.71.min.js:27
When I add the optional configuration {separator: "," , delimiter: "\x22" } the console returns:
Uncaught TypeError: Cannot read property 'rowNum' of undefined jquery.csv-0.71.min.js:27
$.csv.parsers.splitLines jquery.csv-0.71.min.js:27
Original issue: http://code.google.com/p/jquery-csv/issues/detail?id=26
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