Biologia1estradapolimodalpdf34

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Biologia1estradapolimodalpdf34

Update:
The problem in this case is the presence of the. (dot) in the search term. I used the following regex.*?. I notice the match is after the 4th letter of the word:
$pattern = ‘/^.+?(?:“(.+)”|“(.*)”/s;
preg_match_all($pattern, $raw, $matches);

In this case of the HTML scraping the string came in as
「1″) $3en · ” ” $3elf ” ” $3enlis!. $2en.2e $3en · ” ” $3enlis!. $2en.” for
.., the search string was sufficient I guess.

Q:

how to use if statement in bash script

#!/bin/bash

start_number=$1
stop_number=$2
end_number=$3

while true
do
if [ $start_number!= $end_number ]
then
/Applications/XAMPP/xamppfiles/bin/php -f /home/ben/XAMPP/htdocs/sample/run.php $start_number
echo “hi”
fi
sleep 0.5
done

I want to run a script with the user entering start and end number. But if start number is greater than end number the script should start and stop at the respective numbers.
for eg I want to execute the script like this.
./script.sh 10 20 30

and the script should stop at the respective number.

A:

There is no need for while statement.
#!/bin/bash

start_number=$1
stop_number=$2
end_number=$3

if [ $start_number == $end_number ]
then
/Applications/XAMPP/xamppfiles/bin/php -f /home/ben/XAMPP/htdocs/sample/run.php $start_number
echo

1

5
April 1, 2019
df76b833ed
I didn’t get a pass in my first year of school (which I love) but when I did I got something called ‘honor roll’ and a bit of recognition as a ‘rising star’. All because I was the best at something (not sure what was voted on…)
Benjtb702018.
Oct 22, 2017
March 25, 2022 at 5:37 am.
df76b833ed
Passing began there
April 1, 2019
October 14, 2017
Benjtb702018
June 7, 2019
September 27, 2019
-bradman.
April 1, 2019
July 5, 2019
April 1, 2019
September 7, 2019
April 1, 2019
-sorriso.
1

0

0

0
3da54e8ca3

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