I found a few questions here similar to this but they are not quite identical.
I want to grep for the text 'true' format in certain files.
How do I do it?
Not sure how to escape the single quotes around true.
Just wrap the string within double quotes:
grep "'true' format" your_file
Test
$ cat a
this is 'true' format
and this is true formatting
test
$ grep "'true' format" a
this is 'true' format
If you can use the -P (Perl regex syntax) flag of grep, then you can encode ' as \x27 in the pattern:
grep -P '\x27true\x27 format' file.txt
(Similarly, for " use \x22)
Use escapes:
grep -i "'true' format" file.txt
Related
I have a file, the content of file has a string like this:
'/ad/e','#'.base64_decode("ZXZhbA==").'($zad)', 'add'
I want to check the file has this string. But when I use grep to check, It always return false. I try some ways:
grep "'/ad/e','#'.base64_decode("ZXZhbA==").'($zad)', 'add'" foo.txt
grep "'/ad/e','#'\.base64_decode\("ZXZhbA\=\="\)\.'\(\$zad\)', 'add'" foo.txt
str="'/ad/e','#'\.base64_decode\("ZXZhbA\=\="\)\.'\(\$zad\)', 'add'"
grep "$str" foo.txt
Can you help me? Maybe, another command line.
This is my case:
while read str; do
if [ ! -z "$str" ]; then
if grep -Fxq "$str" "$file_path"; then
do somthing
fi
fi
done < <(cat /usr/local/caotoc/db.dat)
Thank you so much!
First, you need to make sure the string is quoted properly. This is a bit of an art form, since your string contains both single and double quotes.
One thought would be to use read and a here-document to avoid having to escape anything.
Second, you need to use -F to perform exact string matching instead of more general regular-expression matching.
IFS= read -r str <<'EOF'
'/ad/e','#'.base64_decode("ZXZhbA==").'($zad)', 'add'
EOF
grep -F "$str" foo.txt
Based on the update, you can use a simple loop to read them one at a time.
while IFS= read -r str; do
grep -F "$str" foo.txt
done < /usr/local/caotoc/db.dat
You may be able to simply use the -f option to grep, which will cause grep to output lines from foo.txt that match any line from db.dat.
grep -f /usr/local/caotoc/db.dat -F foo.txt
Instead of trying to workaround regexes, the simplest way is to turn off regular expressions using -F (or --fixed-strings) option, which makes grep act like a simple string search
-F, --fixed-strings PATTERN is a set of newline-separated strings
like this:
grep -F "'/ad/e','#'.base64_decode(\"ZXZhbA==\").'(\$zad)', 'add'" test
Note: because of the shell, you still need to escape:
double quotes
dollar sign or else $zad is evaluated as an environment variable
I am trying to find alphanumeric string including these two characters "/+" with at least 30 characters in length.
I have written this code,
grep "[a-zA-Z0-9\/\+]{30,}" tmp.txt
cat tmp.txt
> array('rWmyiJgKT8sFXCmMr639U4nWxcSvVFEur9hNOOvQwF/tpYRqTk9yWV2xPFBAZwAPRVs/s
ddd73ZEjfy+airfy8DtqIqKI9+dd 6hdd7soJ9iG0sGs/ld5f2GHzockoYHfh
+pAzx/t17Crf0T/2+8+reo+MU39lqCr02sAkcC1k/LzyBvSDEtu9N/9NHicr jA3SvDqg5s44DFlaNZ/8BW37fGEf2rk13S/q68OVVyzac7IT7yE7PIL9XZ/6LsmrY
KEsAmN4i/+ym8be3wwn KWGYaIB908+7W98pI6qao3iaZB
3mh7Y/nZm52hyLa37978f+PyOCqUh0Wfx2PL3vglofi0l
QVrOM1pg+mFLEIC88B706UzL4Pss7ouEo+EsrES+/qJq9Y1e/UGvwefOWSL2TJdt
this does not work, Mainly I wanted to have minimum length of the string to be 30
In the syntax of grep, the repetition braces need to be backslashed.
grep -o '[a-zA-Z0-9/+]\{30,\}' file
If you want to constrain the match to lines containing only matches to this pattern, add line-start and line-ending anchors:
grep '^[a-zA-Z0-9/+]\{30,\}$' file
The -o option in the first command line causes grep to only print the matching part, not the entire matching line.
The repetition operator is not directly supported in Basic Regular Expression syntax. Use grep -E to enable Extended Regular Expression syntax, or backslash the braces.
You can use
grep -e "^[a-zA-Z0-9/+]\{30,\}" tmp.txt
grep -e "^[a-zA-Z0-9/+]\{30,\}" tmp.txt
+pAzx/t17Crf0T/2+8+reo+MU39lqCr02sAkcC1k/LzyBvSDEtu9N/9NHicr jA3SvDqg5s44DFlaNZ/8BW37fGEf2rk13S/q68OVVyzac7IT7yE7PIL9XZ/6LsmrY
3mh7Y/nZm52hyLa37978f+PyOCqUh0Wfx2PL3vglofi0l
QVrOM1pg+mFLEIC88B706UzL4Pss7ouEo+EsrES+/qJq9Y1e/UGvwefOWSL2TJdt
man grep
Read up about the difference between between regular and extended patterns. You need the -E option.
I Want to search variable and replace with its absolute path in file.
setenv ABC /home/xyz
cat file.txt
${ABC}/Test/Folder_1
${ABC}/Test/Folder_2
I want to replace all occurance of the ${ABC} by /home/xyz.
I tried by the below mentioned way, but does not work,
sed -i 's/\$ABC/echo $ABC/g' file.txt
I can do by below mentioned way, but I do not want to do this way.(I have to put so many back slash)
$ echo $ABC | sed -i 's/\$ABC/\/home\/xyz/g' file.txt
Please give me some suggestion for this question.
Thank You.
If you really want to use the value from a variable in your replacement string, you could use
sed "s#\${ABC}#$ABC#g" file.txt
Character after s in sed is the delimiter and it can be any one character of your choice and it works as long as it's not in the string-to-be-matched and string-to-be-replaced.
Example :
sed 's:string-to-be-matched:string-to-be-replaced:g' file-to-be-edited
: is the delimiter
g means global replacement.
In your case, as the string-to-be-replaced contains the / , the same you are using as sed delimiter.
Simple Solution will be :
sed -i 's:${ABC}:'"$ABC"':g' fill.txt
'" is at either end of $ABC in the replacement string. Purpose is to expand shell variable to use with sed
Another way to get the absolute path is readlink -f ${ABC}/Test/Folder_2
Or the perl alternative to your slash hungry command
$ echo $ABC | sed -i 's/\$ABC/\/home\/xyz/g' file.txt
would be
$ echo $ABC | perl -p -i -e 's!\$ABC!/home/xyz!g'
the first character after the 's above will be used as the delimiter in the replacement expression (i.e. 's#foo#bar#g')
I saw someone use an expression like: sed -e 's, *$,,'
does anybody know why we can use it like this, and what does it do?
I thought the s command should be sed -e 'addr,addrs/reg/sub/' ?
From Using different delimiters in sed:
sed takes whatever follows the "s" as the separator
It is a good way to avoid escaping too much. Code is more readable if you use a delimiter that is not present in the string you want to handle.
For example let's say we want to replace lo/bye from a string. With / as delimiter it would be a little messy:
$ echo "hello/bye" | sed 's/lo\/bye/aa/g'
helaa
So if we define another separator it is more clear:
$ echo "hello/bye" | sed 's|lo/bye|aa|g'
helaa
$ echo "hello/bye" | sed 's,lo/bye,aa,g'
helaa
I need to replace within a little bash script a string inside a file but... I am getting weird results.
Let's say I want to replace:
<tag><![CDATA[text]]></tag>
With:
<tag><![CDATA[replaced_text]]></tag>
Should I use sed? I think due to / and [ ] I am getting weird results...
What would be the best way of approaching this?
Perl with -p option works almost as sed and it has \Q (quote) switch for its regexes:
perl -pe 's{\Q<tag><![CDATA[text]]></tag>}
{<tag><![CDATA[replaced_text]]></tag>}' YOUR_FILE
And in Perl you can use different punctuation to delimiter your expressions (s{...}{...} in my example).
Yes, you need to escape the brackets, and either escape slashes or use different delimiters.
sed 's,<tag><!\[CDATA\[text\]\]></tag>,<tag><!\[CDATA\[replaced)text\]\]></tag>,'
That said, SGML and XML are not actually any better than HTML when it comes to using regexes; don't expect this to generalize.
This should be enough:
$ echo '<tag><![CDATA[text]]></tag>' | sed 's/\[text\]/\[replaced_text\]/'
<tag><![CDATA[replaced_text]]></tag>
You can also change your / separator inside sed to a different character like ,, | or %.
Just use a delimiter other than /, here I use #:
sed -i 's#<tag><!\[CDATA\[text\]\]></tag>#<tag><![CDATA[replaced_text]]></tag>#g' filename
-i to have sed change the file instead of printing it out.
g is for matching more than once (global).
But do you know the exact string you want to match, both the tag and the text?
For instance, if you want to replace the text in all with your replaced_text:
perl -i -pe 's#(<tag><!\[CDATA\[)(.*?)(\]\]></tag>)#\1replaced_text\3#g' filename
Switched to perl because sed doesn't support non-greedy multipliers (the *?).