Re: [Help-bash] Bash read specific element from csv

[Phil]

Hello,

The objective is simply for filenaming purposes.   The file of which a 
snapshot is given underneath comes in everyday.   When the file is 
received it is always called Euronext.csv.   The objective is simply to 
read the data stamp in the third row first column (28 Sep 2014), and 
rename the file from Euronext.csv to Euronext 28 Sep 2014.csv or even 
better Euronext 20140928.csv

Indeed a simple question, but unfortunately one I have not found the 
answer to just yet.
Best regards,

Phil

On 2014-10-07 14:09:59 +0000, Greg Wooledge said:

> On Tue, Oct 07, 2014 at 02:11:55PM +0200, Phil wrote:
> > Hello,
> >
> > I am struggling to find the bash code to read a specific element from a
> > CSV.   The csv is constructed as follows:
> > <--------- SNAP --------->
> > "Naam";"ISIN";"Symbol";"Market";"Trading
> > Currency";"Open";"Hoog";"Laag";"Last";"Last Date/Time";"Time
> > Zone";"Volume";"Turnover"
> > "European Equities";"";;;;;;;;;;;
> > "28 Sep 2014";;;;;;;;;;;;
> > "All datapoints provided as of end of last active trading day.";;;;;;;;;;;;
> > "IDI";"FR0000051393";"IDIP";"Euronext
> > Paris";"EUR";"23.42";"23.50";"23.42";"23.42";"26/09/2014
> > 12:00";"CET";"21";"492.98"
> > "BETER BED";"NL0000339703";"BBED";"Euronext
> > Amsterdam";"EUR";"16.80";"16.80";"16.48";"16.645";"26/09/2014
> > 17:35";"CET";"1206";"20106.64"
> > "ALCATEL-LUCENT";"FR0000130007";"ALU";"Euronext
> > Paris";"EUR";"2.494";"2.504";"2.446";"2.486";"26/09/2014
> > 17:39";"CET";"15581702";"38683091.21"
> > "FIPP";"FR0000038184";"FIPP";"Euronext
> > Paris";"EUR";"0.08";"0.08";"0.08";"0.08";"26/09/2014
> > 10:35";"CET";"20001";"1600.08"
> > "ROBERTET CI";"FR0000045601";"CBE";"Euronext
> > Paris";"EUR";"101.70";"101.70";"101.70";"101.70";"23/09/2014
> > 12:00";"CET";"40";"4068.00"
> > ...
> > <--------- SNAP --------->
>
> Your email program (or your text editor) has mangled the format somewhat.
> I can't tell how many fields there are supposed to be per line.
>
> But given the rest of the question, I guess that doesn't matter:
>
> > I want bash to read the first element on the third line, containing the
> > date (ie 28 Sep 2014), and return into a string.
>
> And then what?  What's the rest of the question?  This is too easy.
> There has to be more.
>
> { read; read; IFS=';' read date _; } < yourfile
> date=${date//\"/}   # to strip the quotation marks
>
> Now that I have answered this question, I *know* the question is going
> to change, and I wish you had asked it all at once, so we can start over
> and do what you really want.
>
> (Once you change the question, this answer will become useless, and
> you'll just throw it away, and take a *completely* different approach to
> parsing the data file.  And then the actual layout of the file is going
> to matter, so we'll have to reverse engineer what the file actually
> looks like before the mangling, etc.)