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Writing array to file using _writeCSVFile()

rajeevanandrajeevanand Members
edited November -1 in Sahi - Open Source
I'm trying to write the contents of my array to a flat file. However the output is appearing in the file like :
"Q","A"," ","S","I","T","E"," ","8","5","2","4",",","Q","A"," ","C","O","U","R","S","E"," ","8","5","2","4""Q","A"," ","S","I","T","E"," ","2","2","9","8",",","Q","A"," ","C","O","U","R","S","E"," ","2","2","9","8""Q","A"," ","S","I","T","E"," ","6","6","9","2",",","Q","A"," ","C","O","U","R","S","E"," ","6","6","9","2""Q","A"," ","S","I","T","E"," ","1","8","8","3",",","Q","A"," ","C","O","U","R","S","E"," ","1","8","8","3"

instead of :

QA SITE 8524,QA COURSE 8524
QA SITE 2298,QA COURSE 2298
.....

Here's my code ($userCountBegin is a random number):
var $arrWriteSiteCourse = new Array();

for ($i = 0; $i < $noOfUsers; $i++)
{
var $site = "QA SITE" + " " + $userCountBegin;
createSite($site);

// Create course

var $course = "QA COURSE" + " " + $userCountBegin;
createCourse($site,$course);

$arrWriteSiteCourse[$i] = $site + "," + $course;
$userCountBegin++;
}

//_alert($arrWriteSiteCourse[0]);
_writeCSVFile($arrWriteSiteCourse,"C:/sahi/userdata/scripts/sitesCourse.csv");

What am I doing wrong here ?

When I use _alert($arrWriteSiteCourse[0]); for debugging, the alert shows:
QA SITE 8524,QA COURSE 8524

Thanks in advance for your help.

Cheers
Rajeev

Comments

  • narayannarayan Administrators
    The data to be passed to _writeCSVFile has to be a 2 dimensional array.

    Change

    $arrWriteSiteCourse[$i] = $site + "," + $course;

    to

    $arrWriteSiteCourse[$i] = [$site, $course];

    and it will work.

    Explanation:

    1) When you set $arrWriteSiteCourse[$i] = $site + "," + $course, Sahi got a string where it was expecting an Array. In javascript, you can index through a string to get each character, which is what Sahi got when it accidentally iterated over the string instead of an array.

    2)

    var $x = is short for

    var $x = new Array();
    $x[0] = 'a';
    $x[1] = 'b';


    Regards,
    Narayan
  • rajeevanandrajeevanand Members
    That worked. Thanks again for your support Narayan.
This discussion has been closed.