Records of each version in list
0
I have a list of versions
1.0.0.1 - 10
1.1.0.1 - 10
1.2.0.1 - 10
That is 30 nr in my list. But I only want to show the 5 highest nr of each sort:
1.0.0.5 - 10
1.1.0.5 - 10
1.2.0.5 - 10
How can I do that? The last nr can be any number but the 3 first nr is only
1.0.0
1.1.0
1.2.0
CODE:
import groovy.json.JsonSlurperClassic
def data = new URL("http://xxxx.se:8081/service/rest/beta/components?repository=Releases").getText()
/**
* 'jsonString' is the input json you have shown
* parse it and store it in collection
*/
Map convertedJSONMap = new JsonSlurperClassic().parseText(data)
def list = convertedJSONMap.items.version
list
groovy
asked Nov 15 '18 at 11:40
Mikael FlobergMikael Floberg
6519
add a comment |
0
I have a list of versions
1.0.0.1 - 10
1.1.0.1 - 10
1.2.0.1 - 10
That is 30 nr in my list. But I only want to show the 5 highest nr of each sort:
1.0.0.5 - 10
1.1.0.5 - 10
1.2.0.5 - 10
How can I do that? The last nr can be any number but the 3 first nr is only
1.0.0
1.1.0
1.2.0
CODE:
import groovy.json.JsonSlurperClassic
def data = new URL("http://xxxx.se:8081/service/rest/beta/components?repository=Releases").getText()
/**
* 'jsonString' is the input json you have shown
* parse it and store it in collection
*/
Map convertedJSONMap = new JsonSlurperClassic().parseText(data)
def list = convertedJSONMap.items.version
list
groovy
asked Nov 15 '18 at 11:40
Mikael FlobergMikael Floberg
6519
add a comment |
0
0
0
I have a list of versions
1.0.0.1 - 10
1.1.0.1 - 10
1.2.0.1 - 10
That is 30 nr in my list. But I only want to show the 5 highest nr of each sort:
1.0.0.5 - 10
1.1.0.5 - 10
1.2.0.5 - 10
How can I do that? The last nr can be any number but the 3 first nr is only
1.0.0
1.1.0
1.2.0
CODE:
import groovy.json.JsonSlurperClassic
def data = new URL("http://xxxx.se:8081/service/rest/beta/components?repository=Releases").getText()
/**
* 'jsonString' is the input json you have shown
* parse it and store it in collection
*/
Map convertedJSONMap = new JsonSlurperClassic().parseText(data)
def list = convertedJSONMap.items.version
list
groovy
asked Nov 15 '18 at 11:40
Mikael FlobergMikael Floberg
6519
I have a list of versions
1.0.0.1 - 10
1.1.0.1 - 10
1.2.0.1 - 10
That is 30 nr in my list. But I only want to show the 5 highest nr of each sort:
1.0.0.5 - 10
1.1.0.5 - 10
1.2.0.5 - 10
How can I do that? The last nr can be any number but the 3 first nr is only
1.0.0
1.1.0
1.2.0
CODE:
import groovy.json.JsonSlurperClassic
def data = new URL("http://xxxx.se:8081/service/rest/beta/components?repository=Releases").getText()
/**
* 'jsonString' is the input json you have shown
* parse it and store it in collection
*/
Map convertedJSONMap = new JsonSlurperClassic().parseText(data)
def list = convertedJSONMap.items.version
list
groovy
groovy
asked Nov 15 '18 at 11:40
Mikael FlobergMikael Floberg
6519
asked Nov 15 '18 at 11:40
Mikael FlobergMikael Floberg
6519
asked Nov 15 '18 at 11:40
Mikael FlobergMikael Floberg
6519
asked Nov 15 '18 at 11:40
Mikael FlobergMikael Floberg
6519
asked Nov 15 '18 at 11:40
Mikael FlobergMikael Floberg
6519
6519
add a comment |
add a comment |
1 Answer
1
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1
Version numbers alone usually don't make an easy sort. So I'd split them into numbers and work from there. E.g.
def versions = [
"1.0.0.12", "1.1.0.42", "1.2.0.666",
"1.0.0.6", "1.1.0.77", "1.2.0.8",
"1.0.0.23", "1.1.0.5", "1.2.0.5",
]
println(
versions.collect
it.split(/./)*.toInteger() // turn into array of integers
.groupBy
it.take(2) // group by the first two numbers
.collect _, vs ->
vs.sort().last() // sort the arrays and take the last
*.join(".") // piece the numbers back together
)
// => [1.0.0.23, 1.1.0.77, 1.2.0.666]
answered Nov 15 '18 at 19:53
cfrickcfrick
18.4k23653
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
Version numbers alone usually don't make an easy sort. So I'd split them into numbers and work from there. E.g.
def versions = [
"1.0.0.12", "1.1.0.42", "1.2.0.666",
"1.0.0.6", "1.1.0.77", "1.2.0.8",
"1.0.0.23", "1.1.0.5", "1.2.0.5",
]
println(
versions.collect
it.split(/./)*.toInteger() // turn into array of integers
.groupBy
it.take(2) // group by the first two numbers
.collect _, vs ->
vs.sort().last() // sort the arrays and take the last
*.join(".") // piece the numbers back together
)
// => [1.0.0.23, 1.1.0.77, 1.2.0.666]
answered Nov 15 '18 at 19:53
cfrickcfrick
18.4k23653
add a comment |
1
Version numbers alone usually don't make an easy sort. So I'd split them into numbers and work from there. E.g.
def versions = [
"1.0.0.12", "1.1.0.42", "1.2.0.666",
"1.0.0.6", "1.1.0.77", "1.2.0.8",
"1.0.0.23", "1.1.0.5", "1.2.0.5",
]
println(
versions.collect
it.split(/./)*.toInteger() // turn into array of integers
.groupBy
it.take(2) // group by the first two numbers
.collect _, vs ->
vs.sort().last() // sort the arrays and take the last
*.join(".") // piece the numbers back together
)
// => [1.0.0.23, 1.1.0.77, 1.2.0.666]
answered Nov 15 '18 at 19:53
cfrickcfrick
18.4k23653
add a comment |
1
1
1
Version numbers alone usually don't make an easy sort. So I'd split them into numbers and work from there. E.g.
def versions = [
"1.0.0.12", "1.1.0.42", "1.2.0.666",
"1.0.0.6", "1.1.0.77", "1.2.0.8",
"1.0.0.23", "1.1.0.5", "1.2.0.5",
]
println(
versions.collect
it.split(/./)*.toInteger() // turn into array of integers
.groupBy
it.take(2) // group by the first two numbers
.collect _, vs ->
vs.sort().last() // sort the arrays and take the last
*.join(".") // piece the numbers back together
)
// => [1.0.0.23, 1.1.0.77, 1.2.0.666]
answered Nov 15 '18 at 19:53
cfrickcfrick
18.4k23653
Version numbers alone usually don't make an easy sort. So I'd split them into numbers and work from there. E.g.
def versions = [
"1.0.0.12", "1.1.0.42", "1.2.0.666",
"1.0.0.6", "1.1.0.77", "1.2.0.8",
"1.0.0.23", "1.1.0.5", "1.2.0.5",
]
println(
versions.collect
it.split(/./)*.toInteger() // turn into array of integers
.groupBy
it.take(2) // group by the first two numbers
.collect _, vs ->
vs.sort().last() // sort the arrays and take the last
*.join(".") // piece the numbers back together
)
// => [1.0.0.23, 1.1.0.77, 1.2.0.666]
answered Nov 15 '18 at 19:53
cfrickcfrick
18.4k23653
answered Nov 15 '18 at 19:53
cfrickcfrick
18.4k23653
answered Nov 15 '18 at 19:53
cfrickcfrick
18.4k23653
answered Nov 15 '18 at 19:53
cfrickcfrick
18.4k23653
18.4k23653
add a comment |
add a comment |
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