regexp for multiline string

[rumburak24]

Hello I am new to reg. exp. in Linux. I try to write a rule for bash.

I have a string like this:

pattern="line 1: Hello, this line is just a report to user....
line2: this line is a separator between two lines
line 3: This line includes pattern. This line was generated in for command."

I would like to get two new string from it:
pattern, including line 3 - without \n - new line mark.
report, including line 1 - without \n - new line mark.

I tried some thing but no result.

Thanks for advice

edit - Solved
$(echo "$pattern" | sed -n '1p')
$(echo "$pattern" | sed -n '3p')

[sunburnt]

Hi rumburak24; If I understand you correctly you want to print out 2 lines without a new line?
2 ways come to my mind:
This outputs the text without a new line: echo -c (text)
Or a list ( more than one line ) can be output without the newlines by: echo $LIST
Having double quotes echos the new lines, like this: echo "$LIST"
But without the double quotes it doesn`t echo the new lines: echo $LIST

If this isn`t what you wanted, please restate your Q...