This is OK
def variables=[
['var1':'test1'],
['var2':'test2'],
['var3':'test3']
]
println "${variables.size()}"
variables.each{entry ->
println "${entry} "
}
I got:
3
[var1:test1]
[var2:test2]
[var3:test3]
but this caused problems
def variables=[
['var1':'test1'],
['var2':'test2'],
['var3':'test3']
]
println "${variables.size()}"
variables.each{entry ->
println "${entry.key} "
}
since I got开发者_运维百科:
3
null
null
null
I'm expecting:
3
var1
var2
var3
what's wrong with my code?
thank you!!!
You want:
def variables=[
'var1':'test1',
'var2':'test2',
'var3':'test3'
]
println variables.size()
variables.each{entry ->
println entry.key
}
Before you had an ArrayList containing three LinkedHashMap objects. The above code is a single LinkedHashMap with three entries. You also don't need string interpolation, so I removed it.
Matthew's solution works great, and it's probably what you wanted (a simpler data structure to begin with).
However, in case you really wanted variables
to be a list of three maps (as per your question), then this how you could get your desired output:
def variables=[
['var1':'test1'],
['var2':'test2'],
['var3':'test3']
]
println "${variables.size()}"
variables.each{ entry->
entry.each {
println it.key
}
}
In the outer closure, every entry
is a map. So we iterate through each of those maps using the inner closure. In this inner closure, every it
closure-param is a key:value pair, so we just print its key using it.key
.
Like Matthew, I've also removed the string interpolation since you don't need it.
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