[SOLVED] Replace third line of nth file with nth line of a single file

Issue

Say I have hundreds of *.xml in /train/xml/, in the following format

# this is the content of /train/xml/RIGHT_NAME.xml
<annotation>
    <path>/train/img/WRONG_NAME.jpg</path>    # this is the WRONG_NAME
</annotation>

The file name WRONG_NAME in <path>...</path> should match that of the .xml file, so that it looks like this:

# this is the content of /train/xml/RIGHT_NAME.xml
<annotation>
    <path>/train/img/RIGHT_NAME.jpg</path>    # this is the **RIGHT_NAME**
</annotation>

One solution I can think of is to:

1. export all file names into a text file:

ls -1 *.xml > filenames.txt

which generates a file with the content:

RIGHT_NAME_0.xml
RIGHT_NAME_1.xml
...

2. then edit filenames.txt, so that it becomes:

# tab at beginning of each line
    <path>/train/img/RIGHT_NAME_0.jpg</path>
    <path>/train/img/RIGHT_NAME_1.jpg</path>
    ...

3. Then, replace the third line of nth .xml file with the nth line from filenames.txt.

Thus the question title.

I've hammered around with sedand awk but had no success. How should I do it (on a EDIT: MacOS machine)? Also, is there a more elegant solution?

Thanks in advance for helping out!

---things I've tried (and didnt work out)---

# this replaces the fifth line with an empty string
for i in *.xml ; do perl -i.bak -pe 's/.*/$i/ if $.==5' RIGHT_NAME.xml ; done

# this apprehends contents of filenames.txt after third line
sed -i.bak -e '/\<path\>/r filenames.txt' RIGHT_NAME.xml

# also, trying to utilize the <path>...</path> pattern...

Solution

Untested:

for xml in *.xml; do
    sed -E -i.bak '3s/([^/]*.jpg)/'"${xml/.xml/.jpg}/" "$xml"
done

Answered By - rici