Question

您好我正在寻找一个解决方案，它将从给定索引的字符串返回一个子字符串。为了避免当前使用if和else check的索引绑定异常。这是一个更好的方法（功能）。

String ordenObtenida = "{\"status\":\"ok\",\"tipo\":\"" + tipo + "\",\"ordenes\":\[;
    while (rs.next()) {
        hasRow = true;
        if (rs.getString("razon") != null) {
            razon = rs.getString("razon").replaceAll("\"", "").trim();
        }
        ordenObtenida += "{\"numero\":\"" + rs.getInt("numero") + "\",\"fecha\":\"" + rs.getDate("fecha") + "\",\"proveedor\":\"" + rs.getInt("proveedor") + "\","
                + "\"codigo\":\"" + rs.getInt("codigo") + "\",\"orden\":\"" + rs.getInt("orden") + "\",\"fepago\":\"" + rs.getDate("fepago") + "\",\"marca\":\"" + rs.getInt("marca") + "\","
                + "\"razon\":\"" + razon + "\",\"importe\":\"" + rs.getFloat("importe") + "\"},";
    }
    ordenObtenida = ordenObtenida.substring(0, ordenObtenida.length() - 1) + "]\"}";

Answer 1

如果索引超出范围，不确定您希望函数做什么，但slice可能符合您的需求：

input.slice(start, end)

一些例子：

scala> "hello".slice(1, 2)
res6: String = e

scala> "hello".slice(1, 30)
res7: String = ello

scala> "hello".slice(7, 8)
res8: String = ""

scala> "hello".slice(0, 5)
res9: String = hello

Answer 2

Try是这样做的一种方式。另一种方法是仅当长度大于使用Option[String]的结束时才应用子字符串。

无效的结束索引

scala> val start = 1
start: Int = 1

scala> val end = 1000
end: Int = 1000

scala> Option("urayagppd").filter(_.length > end).map(_.substring(start, end))
res9: Option[String] = None

有效的结束索引

scala> val end = 6
end: Int = 6

scala> Option("urayagppd").filter(_.length > end).map(_.substring(start, end))
res10: Option[String] = Some(rayag)

此外，您可以将filter和map合并到.collect，如下所示

scala> Option("urayagppd").collect { case x if x.length > end => x.substring(start, end) }
res14: Option[String] = Some(rayag)

scala> val end = 1000
end: Int = 1000

scala> Option("urayagppd").collect { case x if x.length > end => x.substring(start, end) }
res15: Option[String] = None

Scala subString函数

2 个答案: