Bash用变量替换字符串中的第N个单词

Question

I need to find Nth word in a string (space delimited) and replace with a variable. In the below example 4th word needs to be replaced with another string.

1 Test 123456 REPLACE_ME 99

to

1 Test 123456 $STRING_TO_REPLACE 99

I was able to find 4th word using awk '{ print $4}' , but don't know how to replace with another string variable.

Any help will be much appreciated.

Answer 1

   replace='replace me'; echo "233131 2 saad four five dssd sdad" |  awk -v r="$replace" '{ for(i = 1; i <= NF; i++) { if ( i == 4 )print r;else print $i } }'

Answer 2

using awk

str="1 Test 123456 REPLACE_ME 99"
STRING_TO_REPLACE="Hello"

echo $str |awk -v r=${STRING_TO_REPLACE} '{$4=r}1'

Answer 3

Here are two shell solutions. The first is a pure sh solution:

set -- 1 Test 123456 REPLACE_ME 99
one=$1
two=$2
three=$3
shift 4
echo $one $two $three '$STRING_TO_REPLACE' $*

The output is:

1 Test 123456 $STRING_TO_REPLACE 99

The second is a bash solution:

set -- 1 Test 123456 REPLACE_ME 99
echo ${*:1:3} '$STRING_TO_REPLACE' ${*:5:$#}

This outputs:

1 Test 123456 $STRING_TO_REPLACE 99

Answer 4

If you want to replace the N-th word, you could do like this:

awk -v n=4 -v r="$STRING_TO_REPLACE" '{ $n = r } 1' <<< "$str"

Using GNU sed you could replace the 4th word like this:

str="1 Test 123456 REPLACE_ME 99"
STRING_TO_REPLACE="Hello"
sed -e "s/\<\w\+/STRING_TO_REPLACE/4" <<< "$str"

The pattern there means:

• \\< -- start of word
• \\w -- "word character"
• \\+ -- one or more of the previous match
• \\w\\+ -- one or more word characters
• The 4 in s///4 means to perform the replacement for the 4th occurrence only

Answer 5

Another pure sh (possibly bash) solution

set -- 1 Test 123456 REPLACE_ME 99
set -- "${@:0:4}" SOMETHING_ELSE "${@:5}"

Answer 6

if you have Ruby

# echo "1 Test 123456 REPLACE_ME 99" | ruby -e 'f=gets.split(/\s+/);f[3]="new";puts f.join(" ")'
1 Test 123456 new 99