Delimited data contains line feeds where they shouldn't be


 
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# 8  
Old 03-25-2011
Quote:
Originally Posted by dinjo_jo
No i mean can it handle any no of columns irrespective of datatypes ?
Hi dinjo,

Yes, the code work independently the number of columns, just try:

Code:
echo "  1|short desc|long text|2001-01-01 01:01  |long text|2001-01-01 01:01
    2|short desc| long 
    text |2002-02-02 02:02 
    3|short desc|  long  text  | 2003-03-03 03:03 |long text|2001-01-01 01:01
    4|short desc 
    |  long  text    | 2004-04-04 04:04 "  | awk '
 {$1=$1;l=sprintf(/ *[0-9]+\|/?" "RS:FS) $0;printf gensub(/[ \t]+$|[ \t]+?(\|)[ \t]+?/,"\\1","g",l)}' 
 
1|short desc|long text|2001-01-01 01:01|long text|2001-01-01 01:01 
2|short desc|long text|2002-02-02 02:02 
3|short desc|long text|2003-03-03 03:03|long text|2001-01-01 01:01 
4|short desc|long text|2004-04-04 04:04

Regards
# 9  
Old 03-26-2011
Not sure Eric, your code works for me :-) Maybe your version of awk handles input vars differently to mine, I'm using GNU awk 3.1.7

As for the "1", this simply means {print}
# 10  
Old 03-26-2011
maybe you also should handle the case where the \n is inside the 4rth field ...
So you should not only check NF<4 but also the length of the 4rth field
# 11  
Old 03-26-2011
That is a good point, but I suspect that the time stamp will never have this.
But, this did make me think about multiple "\n" on one line. So this will fix it:
Code:
BEGIN { FS = "|" }
NF<4 {
    if (short=="") {
        short=$0
        next
    }
    else {
        short = short" "$0
        next
    }
} 
NF==4 { 
    if (short != "") {
        print short"\n"$0
        short = ""
    }
    else {
        print $0 
    }
}

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