{"id":3186,"date":"2010-04-16T11:57:03","date_gmt":"2010-04-16T11:57:03","guid":{"rendered":""},"modified":"2026-04-18T22:54:23","modified_gmt":"2026-04-18T22:54:23","slug":"bash-insert-content-of-one-file-into-another-file","status":"publish","type":"post","link":"https:\/\/hellem.org\/wblog\/?p=3186","title":{"rendered":"Bash – Insert content of one file into another file"},"content":{"rendered":"

Had to insert a XML-snippet into another XML-file at a given location in a Bash-script<\/a>. Reading the snippet into a variable is quite easy, just by using cat<\/a>. The trouble was how to find the tag to replace most efficiently, then replace it with the xml-snippet. Learned that reading the file content using stdin\/stdout in a bash-script<\/a> can cause side effects not easy to figure out, so that was no option. Instead I first tried to use sed<\/a>, but sed had problems with the fact that the content of the snippet is XML-code. So instead with good help from Geir<\/a> we ended up using Perl, which made it all very simple – everything done in just 2 lines of code.<\/p>\n

First it searches for <!– END jaas –> then replaces the string with the content of the snippet variable and lt;!– END jaas –>
\n
\nvalue=$(cat jaas_snippet.xml)
\nperl -pi -e \"s#<!--\\sEND\\sjaas\\s-->#${value}\\n\\t\\t<!-- END jaas -->#\" jaas.xml
\n<\/code>
\nComments:<\/p>\n