Bash pass date variable as string

Question

I have a script which creates another script when run like this:

cat > "$installpath""tweets.sh" << ENDOFFILE
#!/bin/bash
source "$installpath"config.sh
cd \$webdir
/usr/local/bin/twint -s "\$search" --limit \$scrapelimit -o \$csvname --csv --database \$dbfile -ho
FILE=\$csvname
NAME=\${FILE%.*}
EXT=\${FILE#*.}
DATE=`\date +%d-%m-%Y-%H-%M`
NEWFILE=\${NAME}_\${DATE}.\${EXT}
echo \$NEWFILE
mv \$csvname \$NEWFILE
export NEWFILE
export DATE
ENDOFFILE

However, the script interprets the

DATE=`\date +%d-%m-%Y-%H-%M`

and changes it to

DATE=28-09-2019-15-49

I have tried escaping the variables in every possible way but nothing seems to work. Any ideas?

@Cyrus I have updated the question to include the full thing. Tanks for the heads up! — Niclas, Sep 28 '19 at 15:13

score 0 · Accepted Answer · answered Sep 28 '19 at 15:15

0

I suggest to use:

DATE=\$(date +%d-%m-%Y-%H-%M)

answered Sep 28 '19 at 15:15

Cyrus

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Bash pass date variable as string

1 Answers1