CODEWITHSUNDEEP
bashif-statement
aLee788
aLee788

Reputation: 101

Bash Script - Not returning column/variable

I am having an issue that i cant seem to figure out after a couple hours of tinkering with this. I cant seem this script to return anything to the final column.

#!/bin/bash
file="/Users/USER12/Desktop/url-list.txt"
log="/Users/USER12/Desktop/url-results.txt"
fmt="%-25s%-12s%-16s%-20s\n"
printf "$fmt" DOMAIN_NAME HTTP_CODE RESPONSE_TIME CONTENT_CHECK > "$log"
while read line
do
    read code time < <(curl -o /dev/null --silent --head --write-out '%{http_code} %{time_total}' "$line")
    curl "$line" 2>/dev/null > /Users/USER12/Desktop/domainQueryString_output.txt
    ifStatementConditional=`grep "THE CONTENT I'M LOOKING TO VERIFY" /Users/USER12/Desktop/domainQueryString_output.txt | wc -l`
    if [ $ifStatementConditional -eq 1 ] ; then 
            second_check="online"
        else 
            second_check="DOMAIN IS OFFLINE"
    fi
printf "$fmt" "$line" "$code" "$time" "$second_chance" >> "$log"
done <"$file"

It returns the following but nothing to the final column.... 

DOMAIN_NAME              HTTP_CODE   RESPONSE_TIME   CONTENT_CHECK       
google.com               301         1.177

Thanks for the help guys. The Help is Much Appreciated.

Upvotes: 0

Views: 32

Answers (1)

Petr Skocik
Petr Skocik

Reputation: 60107

You have "$second_chance" where you should have had "$second_check". Other then that, the following is a better way to do your if check:

if grep "THE CONTENT I'M LOOKING TO VERIFY" $yourfile -q
then
  ...
else
  ...
fi

Upvotes: 2

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