使用 grep 提取单引号之间的路径

Question

I'm using wget to download files and during the process, I save log messages (see below) for later use. The most important part is this line Saving to: '/path/somefile.gz' .

I figured out, how I can extract this snipped using grep Saving .
Now, my question is: how can I extract just the path between the single quotes? '/path/somefile.gz' => /path/somefile.gz

HTTP request sent, awaiting response... 200 OK
Length: 15391 (15K) [application/octet-stream]
Saving to: ‘/path/somefile.gz’

     0K .......... .....                                      100% 79,7M=0s

2020-07-06  - ‘/path/somefile.gz’ saved [15391/15391]


Total wall clock time: 0,1s
Downloaded: 1 files, 15K in 0s (79,7 MB/s)

EDIT

Is there any way to process it already in this form?

wget -m --no-parent -nd https://someurl/somefile.gz -P ~/src/  2>&1 |
grep Saving |
tee ~/src/log.txt 

Thank you in advance!

Answer 1

Sample output from wget :

$ cat wget.out
HTTP request sent, awaiting response... 200 OK
Length: 15391 (15K) [application/octet-stream]
Saving to: '/path/somefile.gz'

     0K .......... .....                                      100% 79,7M=0s

2020-07-06  - 'path/somefile.gz' saved [15391/15391]


Total wall clock time: 0,1s
Downloaded: 1 files, 15K in 0s (79,7 MB/s)

One awk solution to extract the desired path/file:

$ awk -F"'" '                        # define input delimiter as single quote
/Saving to:/   { print $2 }          # if line contains string "Saving to:" then print 2nd input field
' wget.out                           # our input
/path/somefile.gz                    # our output

To save the above to a variable:

$ wget_path=$(awk -F"'" '/Saving to:/ {print $2}' wget.out)
$ echo "${wget_path}"
/path/somefile.gz

---

Following up on OP's edit to the question... piping the output of wget into the awk solution:

wget -m --no-parent -nd https://someurl/somefile.gz -P ~/src/ 2>&1 | awk -F"'" '/Saving to:/ {print $2}' | tee ~/src/log.txt 

Answer 2

Since the question asks for a solution in grep , a single GNU grep command to extract the specified path could be:

grep -Po "^Saving to: .\\K[^']*"

provided the Perl Regular Expressions are implemented in the grep (not all grep s implement those).

Of course, it can be used in a pipe also:

wget_command | grep -Po "^Saving to: .\\K[^']*" | tee log.txt

Note that I used a single quote ( ' ) character to anchor the end of path in the pattern match expression, but in the question, Unicode Character Left Single Quotation Mark (U+2018) ( ' ) and Unicode Character Right Single Quotation Mark (U+2019) ( ' ) are used in the sample input. If this is really intended then just replace the [^'] with the [^'] in the pattern match expression above.