How do I split a string into two strings?
1
I already know how to put h4 h7 h8 h9 h7 into an array with space as delimiter, but after that I am trying to separate the h from the numbers to compare the numbers to each other.
How can I do split these in a bash script?
bash string
edited Dec 19 '18 at 11:29
GAD3R
26.1k1751107
asked Dec 19 '18 at 9:48
MercyfonMercyfon
205
add a comment |
1
I already know how to put h4 h7 h8 h9 h7 into an array with space as delimiter, but after that I am trying to separate the h from the numbers to compare the numbers to each other.
How can I do split these in a bash script?
bash string
edited Dec 19 '18 at 11:29
GAD3R
26.1k1751107
asked Dec 19 '18 at 9:48
MercyfonMercyfon
205
add a comment |
1
1
1
1
I already know how to put h4 h7 h8 h9 h7 into an array with space as delimiter, but after that I am trying to separate the h from the numbers to compare the numbers to each other.
How can I do split these in a bash script?
bash string
edited Dec 19 '18 at 11:29
GAD3R
26.1k1751107
asked Dec 19 '18 at 9:48
MercyfonMercyfon
205
I already know how to put h4 h7 h8 h9 h7 into an array with space as delimiter, but after that I am trying to separate the h from the numbers to compare the numbers to each other.
How can I do split these in a bash script?
bash string
bash string
edited Dec 19 '18 at 11:29
GAD3R
26.1k1751107
asked Dec 19 '18 at 9:48
MercyfonMercyfon
205
edited Dec 19 '18 at 11:29
GAD3R
26.1k1751107
asked Dec 19 '18 at 9:48
MercyfonMercyfon
205
edited Dec 19 '18 at 11:29
GAD3R
26.1k1751107
edited Dec 19 '18 at 11:29
GAD3R
26.1k1751107
edited Dec 19 '18 at 11:29
GAD3R
26.1k1751107
26.1k1751107
asked Dec 19 '18 at 9:48
MercyfonMercyfon
205
asked Dec 19 '18 at 9:48
MercyfonMercyfon
205
asked Dec 19 '18 at 9:48
MercyfonMercyfon
205
205
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
4
#!/bin/bash
arr=( h4 h7 h8 h9 h7 )
for thing in "${arr[@]}"; do
num=${thing#?}
printf 'The number in "%s" is %dn' "$thing" "$num"
done
The variable expansion ${variable#pattern} removes the shortest prefix string matching pattern from $variable. The pattern ? matches a single character.
The output will be
The number in "h4" is 4
The number in "h7" is 7
The number in "h8" is 8
The number in "h9" is 9
The number in "h7" is 7
Alternatively, to ignore remove the non-digit prefix regardless of how long it is, using sed,
#!/bin/bash
arr=( hello4 happy7 hobbit8 hulk9 hr-department7 )
for thing in "${arr[@]}"; do
num=$( printf '%sn' "$thing" | sed 's/^[^[:digit:]]*//' )
printf 'The number in "%s" is %dn' "$thing" "$num"
done
or, using regular expression matching of the digits at the end,
#!/bin/bash
arr=( hello4 happy7 hobbit8 hulk9 hr-department7 )
for thing in "${arr[@]}"; do
if [[ "$thing" =~ ([[:digit:]]+)$ ]]; then
printf 'The number in "%s" is %dn' "$thing" "${BASH_REMATCH[1]}"
fi
done
answered Dec 19 '18 at 9:56
KusalanandaKusalananda
125k16236389
+1 Nice, Can you please explain this part:printf 'The number in "%s" is %dn' "$thing" "$num"?
– User123
Dec 19 '18 at 10:02
2
@User123 It's a standard way of outputting a string containing variable data.printftakes a format string containing format specifiers (here%sfor a string and%dfor a decimal number). The variable data is provided as the extra arguments, andprintfwill insert the data in the appropriate places in the format string. It works pretty much the same way as in most programming languages (C and Perl, for example). See also Why is printf better than echo?
– Kusalananda
Dec 19 '18 at 10:04
Thanks!! But if i use:echo "The number in $thing is $num"It'll also produce the same output. Is there any downfall in using like this?
– User123
Dec 19 '18 at 10:08
2
@User123 Not if$thingcontains escape sequences liken. I know it's very unlikely that your data may uset,netc., but other people's data may, and my answer is for anyone who reads it and has a similar problem. Usingechoin their code may do unexpected things depending on what their data actually is.printfguarantees more consistent output. See Why is printf better than echo?
– Kusalananda
Dec 19 '18 at 10:11
@User123 Sorry, I confused you with the person asking the question. Nevertheless...
– Kusalananda
Dec 19 '18 at 10:14
|
show 1 more comment
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1 Answer
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1 Answer
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4
#!/bin/bash
arr=( h4 h7 h8 h9 h7 )
for thing in "${arr[@]}"; do
num=${thing#?}
printf 'The number in "%s" is %dn' "$thing" "$num"
done
The variable expansion ${variable#pattern} removes the shortest prefix string matching pattern from $variable. The pattern ? matches a single character.
The output will be
The number in "h4" is 4
The number in "h7" is 7
The number in "h8" is 8
The number in "h9" is 9
The number in "h7" is 7
Alternatively, to ignore remove the non-digit prefix regardless of how long it is, using sed,
#!/bin/bash
arr=( hello4 happy7 hobbit8 hulk9 hr-department7 )
for thing in "${arr[@]}"; do
num=$( printf '%sn' "$thing" | sed 's/^[^[:digit:]]*//' )
printf 'The number in "%s" is %dn' "$thing" "$num"
done
or, using regular expression matching of the digits at the end,
#!/bin/bash
arr=( hello4 happy7 hobbit8 hulk9 hr-department7 )
for thing in "${arr[@]}"; do
if [[ "$thing" =~ ([[:digit:]]+)$ ]]; then
printf 'The number in "%s" is %dn' "$thing" "${BASH_REMATCH[1]}"
fi
done
answered Dec 19 '18 at 9:56
KusalanandaKusalananda
125k16236389
+1 Nice, Can you please explain this part:printf 'The number in "%s" is %dn' "$thing" "$num"?
– User123
Dec 19 '18 at 10:02
2
@User123 It's a standard way of outputting a string containing variable data.printftakes a format string containing format specifiers (here%sfor a string and%dfor a decimal number). The variable data is provided as the extra arguments, andprintfwill insert the data in the appropriate places in the format string. It works pretty much the same way as in most programming languages (C and Perl, for example). See also Why is printf better than echo?
– Kusalananda
Dec 19 '18 at 10:04
Thanks!! But if i use:echo "The number in $thing is $num"It'll also produce the same output. Is there any downfall in using like this?
– User123
Dec 19 '18 at 10:08
2
@User123 Not if$thingcontains escape sequences liken. I know it's very unlikely that your data may uset,netc., but other people's data may, and my answer is for anyone who reads it and has a similar problem. Usingechoin their code may do unexpected things depending on what their data actually is.printfguarantees more consistent output. See Why is printf better than echo?
– Kusalananda
Dec 19 '18 at 10:11
@User123 Sorry, I confused you with the person asking the question. Nevertheless...
– Kusalananda
Dec 19 '18 at 10:14
|
show 1 more comment
4
#!/bin/bash
arr=( h4 h7 h8 h9 h7 )
for thing in "${arr[@]}"; do
num=${thing#?}
printf 'The number in "%s" is %dn' "$thing" "$num"
done
The variable expansion ${variable#pattern} removes the shortest prefix string matching pattern from $variable. The pattern ? matches a single character.
The output will be
The number in "h4" is 4
The number in "h7" is 7
The number in "h8" is 8
The number in "h9" is 9
The number in "h7" is 7
Alternatively, to ignore remove the non-digit prefix regardless of how long it is, using sed,
#!/bin/bash
arr=( hello4 happy7 hobbit8 hulk9 hr-department7 )
for thing in "${arr[@]}"; do
num=$( printf '%sn' "$thing" | sed 's/^[^[:digit:]]*//' )
printf 'The number in "%s" is %dn' "$thing" "$num"
done
or, using regular expression matching of the digits at the end,
#!/bin/bash
arr=( hello4 happy7 hobbit8 hulk9 hr-department7 )
for thing in "${arr[@]}"; do
if [[ "$thing" =~ ([[:digit:]]+)$ ]]; then
printf 'The number in "%s" is %dn' "$thing" "${BASH_REMATCH[1]}"
fi
done
answered Dec 19 '18 at 9:56
KusalanandaKusalananda
125k16236389
+1 Nice, Can you please explain this part:printf 'The number in "%s" is %dn' "$thing" "$num"?
– User123
Dec 19 '18 at 10:02
2
@User123 It's a standard way of outputting a string containing variable data.printftakes a format string containing format specifiers (here%sfor a string and%dfor a decimal number). The variable data is provided as the extra arguments, andprintfwill insert the data in the appropriate places in the format string. It works pretty much the same way as in most programming languages (C and Perl, for example). See also Why is printf better than echo?
– Kusalananda
Dec 19 '18 at 10:04
Thanks!! But if i use:echo "The number in $thing is $num"It'll also produce the same output. Is there any downfall in using like this?
– User123
Dec 19 '18 at 10:08
2
@User123 Not if$thingcontains escape sequences liken. I know it's very unlikely that your data may uset,netc., but other people's data may, and my answer is for anyone who reads it and has a similar problem. Usingechoin their code may do unexpected things depending on what their data actually is.printfguarantees more consistent output. See Why is printf better than echo?
– Kusalananda
Dec 19 '18 at 10:11
@User123 Sorry, I confused you with the person asking the question. Nevertheless...
– Kusalananda
Dec 19 '18 at 10:14
|
show 1 more comment
4
4
4
#!/bin/bash
arr=( h4 h7 h8 h9 h7 )
for thing in "${arr[@]}"; do
num=${thing#?}
printf 'The number in "%s" is %dn' "$thing" "$num"
done
The variable expansion ${variable#pattern} removes the shortest prefix string matching pattern from $variable. The pattern ? matches a single character.
The output will be
The number in "h4" is 4
The number in "h7" is 7
The number in "h8" is 8
The number in "h9" is 9
The number in "h7" is 7
Alternatively, to ignore remove the non-digit prefix regardless of how long it is, using sed,
#!/bin/bash
arr=( hello4 happy7 hobbit8 hulk9 hr-department7 )
for thing in "${arr[@]}"; do
num=$( printf '%sn' "$thing" | sed 's/^[^[:digit:]]*//' )
printf 'The number in "%s" is %dn' "$thing" "$num"
done
or, using regular expression matching of the digits at the end,
#!/bin/bash
arr=( hello4 happy7 hobbit8 hulk9 hr-department7 )
for thing in "${arr[@]}"; do
if [[ "$thing" =~ ([[:digit:]]+)$ ]]; then
printf 'The number in "%s" is %dn' "$thing" "${BASH_REMATCH[1]}"
fi
done
answered Dec 19 '18 at 9:56
KusalanandaKusalananda
125k16236389
#!/bin/bash
arr=( h4 h7 h8 h9 h7 )
for thing in "${arr[@]}"; do
num=${thing#?}
printf 'The number in "%s" is %dn' "$thing" "$num"
done
The variable expansion ${variable#pattern} removes the shortest prefix string matching pattern from $variable. The pattern ? matches a single character.
The output will be
The number in "h4" is 4
The number in "h7" is 7
The number in "h8" is 8
The number in "h9" is 9
The number in "h7" is 7
Alternatively, to ignore remove the non-digit prefix regardless of how long it is, using sed,
#!/bin/bash
arr=( hello4 happy7 hobbit8 hulk9 hr-department7 )
for thing in "${arr[@]}"; do
num=$( printf '%sn' "$thing" | sed 's/^[^[:digit:]]*//' )
printf 'The number in "%s" is %dn' "$thing" "$num"
done
or, using regular expression matching of the digits at the end,
#!/bin/bash
arr=( hello4 happy7 hobbit8 hulk9 hr-department7 )
for thing in "${arr[@]}"; do
if [[ "$thing" =~ ([[:digit:]]+)$ ]]; then
printf 'The number in "%s" is %dn' "$thing" "${BASH_REMATCH[1]}"
fi
done
answered Dec 19 '18 at 9:56
KusalanandaKusalananda
125k16236389
edited Dec 19 '18 at 11:41
answered Dec 19 '18 at 9:56
KusalanandaKusalananda
125k16236389
answered Dec 19 '18 at 9:56
KusalanandaKusalananda
125k16236389
answered Dec 19 '18 at 9:56
KusalanandaKusalananda
125k16236389
125k16236389
+1 Nice, Can you please explain this part:printf 'The number in "%s" is %dn' "$thing" "$num"?
– User123
Dec 19 '18 at 10:02
2
@User123 It's a standard way of outputting a string containing variable data.printftakes a format string containing format specifiers (here%sfor a string and%dfor a decimal number). The variable data is provided as the extra arguments, andprintfwill insert the data in the appropriate places in the format string. It works pretty much the same way as in most programming languages (C and Perl, for example). See also Why is printf better than echo?
– Kusalananda
Dec 19 '18 at 10:04
Thanks!! But if i use:echo "The number in $thing is $num"It'll also produce the same output. Is there any downfall in using like this?
– User123
Dec 19 '18 at 10:08
2
@User123 Not if$thingcontains escape sequences liken. I know it's very unlikely that your data may uset,netc., but other people's data may, and my answer is for anyone who reads it and has a similar problem. Usingechoin their code may do unexpected things depending on what their data actually is.printfguarantees more consistent output. See Why is printf better than echo?
– Kusalananda
Dec 19 '18 at 10:11
@User123 Sorry, I confused you with the person asking the question. Nevertheless...
– Kusalananda
Dec 19 '18 at 10:14
|
show 1 more comment
+1 Nice, Can you please explain this part:printf 'The number in "%s" is %dn' "$thing" "$num"?
– User123
Dec 19 '18 at 10:02
2
@User123 It's a standard way of outputting a string containing variable data.printftakes a format string containing format specifiers (here%sfor a string and%dfor a decimal number). The variable data is provided as the extra arguments, andprintfwill insert the data in the appropriate places in the format string. It works pretty much the same way as in most programming languages (C and Perl, for example). See also Why is printf better than echo?
– Kusalananda
Dec 19 '18 at 10:04
Thanks!! But if i use:echo "The number in $thing is $num"It'll also produce the same output. Is there any downfall in using like this?
– User123
Dec 19 '18 at 10:08
2
@User123 Not if$thingcontains escape sequences liken. I know it's very unlikely that your data may uset,netc., but other people's data may, and my answer is for anyone who reads it and has a similar problem. Usingechoin their code may do unexpected things depending on what their data actually is.printfguarantees more consistent output. See Why is printf better than echo?
– Kusalananda
Dec 19 '18 at 10:11
@User123 Sorry, I confused you with the person asking the question. Nevertheless...
– Kusalananda
Dec 19 '18 at 10:14
+1 Nice, Can you please explain this part:
printf 'The number in "%s" is %dn' "$thing" "$num" ?– User123
Dec 19 '18 at 10:02
+1 Nice, Can you please explain this part:
printf 'The number in "%s" is %dn' "$thing" "$num" ?– User123
Dec 19 '18 at 10:02
2
2
@User123 It's a standard way of outputting a string containing variable data.
printf takes a format string containing format specifiers (here %s for a string and %d for a decimal number). The variable data is provided as the extra arguments, and printf will insert the data in the appropriate places in the format string. It works pretty much the same way as in most programming languages (C and Perl, for example). See also Why is printf better than echo?– Kusalananda
Dec 19 '18 at 10:04
@User123 It's a standard way of outputting a string containing variable data.
printf takes a format string containing format specifiers (here %s for a string and %d for a decimal number). The variable data is provided as the extra arguments, and printf will insert the data in the appropriate places in the format string. It works pretty much the same way as in most programming languages (C and Perl, for example). See also Why is printf better than echo?– Kusalananda
Dec 19 '18 at 10:04
Thanks!! But if i use:
echo "The number in $thing is $num" It'll also produce the same output. Is there any downfall in using like this?– User123
Dec 19 '18 at 10:08
Thanks!! But if i use:
echo "The number in $thing is $num" It'll also produce the same output. Is there any downfall in using like this?– User123
Dec 19 '18 at 10:08
2
2
@User123 Not if
$thing contains escape sequences like n. I know it's very unlikely that your data may use t, n etc., but other people's data may, and my answer is for anyone who reads it and has a similar problem. Using echo in their code may do unexpected things depending on what their data actually is. printf guarantees more consistent output. See Why is printf better than echo?– Kusalananda
Dec 19 '18 at 10:11
@User123 Not if
$thing contains escape sequences like n. I know it's very unlikely that your data may use t, n etc., but other people's data may, and my answer is for anyone who reads it and has a similar problem. Using echo in their code may do unexpected things depending on what their data actually is. printf guarantees more consistent output. See Why is printf better than echo?– Kusalananda
Dec 19 '18 at 10:11
@User123 Sorry, I confused you with the person asking the question. Nevertheless...
– Kusalananda
Dec 19 '18 at 10:14
@User123 Sorry, I confused you with the person asking the question. Nevertheless...
– Kusalananda
Dec 19 '18 at 10:14
|
show 1 more comment
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