How to use grep to get anything just after `name=`?

As detailed here, you want a positive lookbehind clause, such as:

grep -P '(?<=name=)[ A-Za-z0-9]*' filename

The -P makes grep use the Perl dialect, otherwise you'd probably need to escape the parentheses. You can also, as noted elsewhere, append the -o parameter to print out only what is matched. The part in brackets specifies that you want alphanumerics and spaces.

The advantage of using a positive lookbehind clause is that the "name=" text is not part of the match. If grep highlights matched text, it will only highlight the alphanumeric (and spaces) part. The -o parameter will also not display the "name=" part. And, if you transition this to another program like sed that might capture the text and do something with it, you won't be capturing the "name=" part, although you can also do that by using capturing parenthess.

Try this:

sed -n 's/^name=//p' filename

It tells sed to print nothing (-n) by default, substitute your prefix with nothing, and print if the substitution occurs.

Bonus: if you really need it to only match entries with only spaces and alphanumerics, you can do that too:

sed -n 's/^name=\([ 0-9a-zA-Z]*$\)/\1/p' filename

Here we've added a pattern to match spaces and alphanumerics only until the end of the line ($), and if we match we substitute the group in parentheses and print.

grep does not extract like you expect. What you need is

grep "name=" file.txt | cut -d'=' -f1-

gawk

echo "name=some value here" | awk -F"=" '/name=/ { print $2}' 

or with bash

str="name=some value here"
IFS="="
set -- $str
echo $1    
unset IFS

or

str="name=some value here"
str=${str/name=/}

grep will print the entire line where it matches the pattern. To print only the pattern matched, use the grep -o option. You'll probably also need to use sed to remove the name= part of the pattern.

grep -o 'name=[0-9a-zA-Z ]'  myfile | sed /^name=/d