how to pass "one" argument and use it twice in "xargs" command
bash, echo, popen, shell, xargs
Solution
If you can't change the input format, you could set the delimiter to a space:
$ echo -n {0..4} | xargs -d " " -I@ echo @,@
0,0
1,1
2,2
3,3
4,4
Otherwise, change the input to separate the tokens with a newline:
$ printf "%s\n" {0..4} | xargs -I@ echo @,@
0,0
1,1
2,2
3,3
4,4
The reason for this syntax is explained in `man xargs`
-I replace-str
Replace occurrences of replace-str in the initial-arguments with names read from
standard input. Also, unquoted blanks do not terminate input items; instead the sep‐
arator is the newline character. Implies -x and -L 1.
So you must set the delimiter manually to a space if you want to delimit fields.
Problem
I tried to use the `xargs` to pass the arguments to the `echo`: ``` [usr@linux scripts]$ echo {0..4} | xargs -n 1 echo 0 1 2 3 4 ``` the `-n 1` insured that the `xargs` pass `1` arguments a time to the `echo`. Then I want to use this aruments twice, however the results is not I wanted: ``` [usr@linux scripts]$ echo {0..4} | xargs -I@ -n 1 echo @,@ 0 1 2 3 4,0 1 2 3 4 ``` the `-n 1` seems disabled when I added the `-I@`, and this is the result I wanted: ``` 0,0 1,1 2,2 3,3 4,4 ``` how can I achieve that? --------Supply------------------ I have used the method recommanded by @123 ,however ,there are still another question: test.sh: ``` #!/bin/bash a[0]=1 a[1]=2 echo "a[0] and a[1] : "${a[0]}, ${a[1]} echo -n {0..1} | xargs -I num -d" " echo num,${a[num]},num ``` and this is the output: ``` [usr@linux scripts]$ sh test.sh a[0] and a[1] : 1, 2 0,1,0 1,1,1 ``` you can see that the array `a` is not returned the value I wanted :< And How can I fix this problem?