我宁愿用这种丑陋的方式从列表中构建字符串:
val input = listOf("[A,B]", "[C,D]")
val builder = StringBuilder()
builder.append("Serialized('IDs((")
for (pt in input) {
builder.append(pt[0] + " " + pt[1])
builder.append(", ")
}
builder.append("))')")
问题是，它在最后一个元素后添加了一个逗号，如果我想避免，我需要在循环中为最后一个元素添加另一个if check.
I wonder if there is a more concise way of doing this in kotlin?
EDIT
最终结果应该是:
Serialized('IDs((A B,C D))')
在Kotlin中，这种用例可以使用joinToString(它只涉及插入分隔符between个元素).
It is very versatile because it allows to specify a transform function for each element (in addition to the more classic separator, prefix, postfix). This makes it equivalent to mapping all elements to strings and then joining them together, but in one single call.
If input really is a List<List<String>> like you mention in the title and you assume in your loop, you can use:
input.joinToString(
prefix = "Serialized('IDs((",
postfix = "))')",
separator = ", ",
) { (x, y) -> "$x $y" }
注意，带(x, y)的语法是一种解构语法，它会自动获取列表中的第一个和第二个元素(括号很重要).
If your input is in fact a List<String> as in listOf("[A,B]", "[C,D]") that you wrote at the top of your code, you can instead use:
input.joinToString(
prefix = "Serialized('IDs((",
postfix = "))')",
separator = ", ",
) { it.removeSurrounding("[", "]").replace(",", " ") }