我试着:
somearray = ["some", "thing"]
anotherarray = ["another", "thing"]
somearray.push(anotherarray.flatten!)
我预期的
["some", "thing", "another", "thing"]
但有
["some", "thing", nil]
当前回答
somearray = ["some", "thing"]
anotherarray = ["another", "thing"]
somearray + anotherarray # => ["some", "thing", "another", "thing"]
somearray.concat anotherarray # => ["some", "thing", "another", "thing"]
somearray.push(anotherarray).flatten # => ["some", "thing", "another", "thing"]
somearray.push *anotherarray # => ["another", "thing", "another", "thing"]
其他回答
这里有两种方法，注意在这个例子中，第一种方法分配一个新数组(转换为somearray = somearray + anotherarray)
somearray = ["some", "thing"]
anotherarray = ["another", "thing"]
somearray += anotherarray # => ["some", "thing", "another", "thing"]
somearray = ["some", "thing"]
somearray.concat anotherarray # => ["some", "thing", "another", "thing"]
试试这个，它将结合你的数组删除重复
array1 = ["foo", "bar"]
array2 = ["foo1", "bar1"]
array3 = array1|array2
http://www.ruby-doc.org/core/classes/Array.html
进一步的文档请参阅“Set Union”
(array1 + array2).uniq
这样你就可以先得到array1元素。你不会得到副本。
["some", "thing"] + ["another", "thing"]
你有一个可行的想法，但#flatten!在错误的地方——它会使接收器变平，所以你可以用它把[1,2，['foo'， 'bar']]变成[1,2，'foo'，'bar']。
毫无疑问，我忘记了一些方法，但你可以串联起来:
a1.concat a2
a1 + a2 # creates a new array, as does a1 += a2
或预先考虑/追加:
a1.push(*a2) # note the asterisk
a2.unshift(*a1) # note the asterisk, and that a2 is the receiver
或接头:
a1[a1.length, 0] = a2
a1[a1.length..0] = a2
a1.insert(a1.length, *a2)
或append和flatten:
(a1 << a2).flatten! # a call to #flatten instead would return a new array
