PHWinfo - Afficher un message - cgi using shell script : unexpected behavior

15/03/2008, 02h26

Hi everyone,
I'm writing a small shell script to act like a cgi
program. I have linked this script to a basic site I have developed
via a form. I have attached the code below, and would greatly
appreciate any pointers to correct glaring mistakes

..

The problem: Everything displays as expected until I hit the second if
statement. When I uncomment this statement all I get is a blank page,
If I comment out this part ( if [ $flag1 -e 1 ]; then ) then I see
what should be displayed until this point. The syntax for the if
statement seems correct... I can't figure out what stupid mistake is
causing this error

To recap: When I comment out the part if [ $flag1 -e 1 ]; then, I can
see Your URL is www.hfdjhgjghjgh.com and then the flag1 value. As soon
as I enable the if part I see a blank page! arrrrgh! please :-)

#!/bin/sh

wd=$PWD

echo "Content-type: text/html"
echo

echo "<HTML><HEAD><TITLE>My first cgi</Title></HEAD>"
echo " <BODY>"
echo " <pre>"

input=(`echo "$QUERY_STRING"`)

#check for ; and |
flag=(`echo $input | awk 'BEGIN{flag=0} /;|%3B/{flag=1} /\||%7C/
{flag=2} END{print flag}'`)

if [ $flag -e 0 ];
then
echo "Malformed URL: invalid characters used"
echo "Please check URL"
else
burl=(`echo $input | awk '{split($0,a,"&");split(a[1],b,"=")

rint
b[2]}'`);
echo "<H3>Your URL is </H3> $burl";
#push this in psql db and check if we know
#about it
echo "SELECT sitename FROM urltable WHERE sitename = '$burl';" >
$wd/query.sql
flag1=(`psql -f $wd/query.sql urldb | awk 'BEGIN{a=0} /(0 rows)/
{a=1} END{print a}'`)
echo $flag1

if [ $flag1 -e 1 ]; then
#PROBLEM IS HERE
else
#do something
fi
#return the formatted page
fi
rm -f $wd/query.sql
echo "</BODY>"
echo "</HTML>"
exit

15/03/2008, 02h26	#1
newbiegalore Messages: n/a Hébergeur:	cgi using shell script : unexpected behavior Hi everyone, I'm writing a small shell script to act like a cgi program. I have linked this script to a basic site I have developed via a form. I have attached the code below, and would greatly appreciate any pointers to correct glaring mistakes .. The problem: Everything displays as expected until I hit the second if statement. When I uncomment this statement all I get is a blank page, If I comment out this part ( if [ $flag1 -e 1 ]; then ) then I see what should be displayed until this point. The syntax for the if statement seems correct... I can't figure out what stupid mistake is causing this error To recap: When I comment out the part if [ $flag1 -e 1 ]; then, I can see Your URL is www.hfdjhgjghjgh.com and then the flag1 value. As soon as I enable the if part I see a blank page! arrrrgh! please :-) #!/bin/sh wd=$PWD echo "Content-type: text/html" echo echo "<HTML><HEAD><TITLE>My first cgi</Title></HEAD>" echo " <BODY>" echo " <pre>" input=(`echo "$QUERY_STRING"`) #check for ; and \| flag=(`echo $input \| awk 'BEGIN{flag=0} /;\|%3B/{flag=1} /\\|\|%7C/ {flag=2} END{print flag}'`) if [ $flag -e 0 ]; then echo "Malformed URL: invalid characters used" echo "Please check URL" else burl=(`echo $input \| awk '{split($0,a,"&");split(a[1],b,"=")rint b[2]}'`); echo "<H3>Your URL is </H3> $burl"; #push this in psql db and check if we know #about it echo "SELECT sitename FROM urltable WHERE sitename = '$burl';" > $wd/query.sql flag1=(`psql -f $wd/query.sql urldb \| awk 'BEGIN{a=0} /(0 rows)/ {a=1} END{print a}'`) echo $flag1 if [ $flag1 -e 1 ]; then #PROBLEM IS HERE else #do something fi #return the formatted page fi rm -f $wd/query.sql echo "</BODY>" echo "</HTML>" exit