Script IF condition problem

Hi experts
data var type i value 10. defined in print program,
i wrote a text element in scrpt form
/E ELE
/: IF &VAR& EQ 10
welcome
/: ELSE
bye
/:ENDIF
called in write_form
Output : bye
why it executes else part, please help me
thanks
sitaram

Hi Change your code to the following.
/E ELE
/: IF &VAR& = '10'
welcome
/: ELSE
bye
/:ENDIF

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Hope it helps,
Adele
<i>PS: I have seen that every post you make you set as answered/solved problem already. Please make sure which problems are solved and which not. People tend to not look at solved posts that often. Every question you asked have been marked as solved before anyone replied. Were they solved?</i>

Bash script - sqllite insert problem

So, I'm making a bash script that inserts a config file into a database, but I can't get it to work.
This is my best attempt so far:
name="bashrc"
location="/home/user/.bashrc"
content=`cat "$location"`
sqlite3 testdb "insert into dbase values ('`echo $name`','`echo "$location"`','`echo "$content"`')"
the table is
create table dbase(name text, location text, data text)
I had a hard time trying to figure out how I'd be able to get $variables to work (because of the quotes) and this echo style seems to work, but there seems to be no way to insert the config file into the database.
plaintext files seem to work no problem, but when I'm trying to insert some config file, it spits out errors. (It seems like it's trying to execute the file somehow?)
Last edited by initbox (2009-08-29 15:28:49)

initbox wrote:I was thinking the single quotes were stopping bash from replacing $variables with the content, hence the echoes.
your single quotes are inside a double quote, so to bash they're just another character in the string.
I'm still getting an SQL syntax error, so bash is no longer getting in the way, but this thing is a true nightmare. What next? This turned out to be way harder than I expected, I'm just doing this as a "fun" learning experiment.
Your error is probably because I was escaping the quotes for bash, and I should have been escaping them for sqlite. In sqlite, escaping a quote is done by doubling it, so ' is escaped to '' (that's two apostrephes side by side, not a double quote. Compare '' and ").
This untested sed command might work:
content = $(cat $location | sed "s/\'/\'\'/g")
The \'s are escaping the quote from bash, what its really doing is replacing ' with ''.
Dusty

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