Delete lines with lower/higher ISO date using awk and/or sed -

i've following file:

data;ora;testo log;ambiente;jobname;nomeapp;nomescript;parametri; 20130216;17-02-2014;15:15:13;devi valorizzare la variabile ambiente con "bcoll" o "bprod" ;; 20130427;17-02-2014;15:15:13;devi valorizzare la variabile ambiente con "bcoll" o "bprod" ;; 20131010;17-02-2014;15:15:13;devi valorizzare la variabile ambiente con "bcoll" o "bprod" ;; 20130217;17-02-2014;15:15:13;devi valorizzare la variabile ambiente con "bcoll" o "bprod" ;; 20130217;17-02-2014;15:15:13;devi valorizzare la variabile ambiente con "bcoll" o "bprod" ;; 20130217;17-02-2014;15:15:13;devi valorizzare la variabile ambiente con "bcoll" o "bprod" ;; 20140217;17-02-2014;15:15:13;devi valorizzare la variabile ambiente con "bcoll" o "bprod" ;; 

the first line date in iso format increasing in every line.

i want cut line older given date using 1 line of shell script grep or awk or sed.

be careful header description fields.

it's important compatible aix, linux, solaris, not sed -i.

i've command extract line first empy line.

awk -f ";" '$1 < 20140101 {print $1}' log.csv 

piping | awk '!/$1/' log.csv doesnt' work maybe because there's multiple lines.

it doesn't work piping sed:

sed '/$1/ d' < log.csv 

but 1 doesn't work:

for in awk -f ";" '$1 < 20140101 {print $1}' log.csv; awk '!/$i/' log.csv; done 

nr==1 true in awk first line. rest, use comparison

awk -f\; 'nr==1 || $1 < 20140101 { print }'