convert String to Option[String]
I am new to scala programming and am facing difficulty to convert from String to Option[String]. I am not sure if the approach I am following is correct or not. Here is the code:
EmpInfo.scala:
package model
case class EmpInfo(empType: Option[String]) extends Named {
override def name: String = clientName
}
EmpReconciliator.scala
import scalaz._
import Scalaz._
import scala.concurrent.Future
import play.api.mvc.Action
import play.api.mvc.AnyContent
import model.EmpInfo
trait EmpReconciliator[A] extends BaseController[A] {
val oldName = "abc"
val newName = "xyz"
def reconcileGeneric(genericResource: A) = {
(genericResource match {
case empInfo : EmpInfo => reconcileEmpInfo(empInfo)
case other@_ => other
}).asInstanceOf[A]
}
def reconcileEmpInfo(empInfo: EmpInfo) = {
empInfo.empTypeType match {
case Some(oldName) => empInfo.copy(empType = Some(newName))
case _ => empInfo
}
}
}
Here, I need to replace all instances of oldName with newName.
But instead, I am getting "1" in the response. I cannot change type of empType: Option[String] to String (need it that way).
I know it's a very minor change needed but cannot place what.
Please help.
scala playframework
edited May 22 '17 at 9:24
Jasper-M
9,09211329
asked May 22 '17 at 8:57
b1399877b1399877
125
add a comment |
I am new to scala programming and am facing difficulty to convert from String to Option[String]. I am not sure if the approach I am following is correct or not. Here is the code:
EmpInfo.scala:
package model
case class EmpInfo(empType: Option[String]) extends Named {
override def name: String = clientName
}
EmpReconciliator.scala
import scalaz._
import Scalaz._
import scala.concurrent.Future
import play.api.mvc.Action
import play.api.mvc.AnyContent
import model.EmpInfo
trait EmpReconciliator[A] extends BaseController[A] {
val oldName = "abc"
val newName = "xyz"
def reconcileGeneric(genericResource: A) = {
(genericResource match {
case empInfo : EmpInfo => reconcileEmpInfo(empInfo)
case other@_ => other
}).asInstanceOf[A]
}
def reconcileEmpInfo(empInfo: EmpInfo) = {
empInfo.empTypeType match {
case Some(oldName) => empInfo.copy(empType = Some(newName))
case _ => empInfo
}
}
}
Here, I need to replace all instances of oldName with newName.
But instead, I am getting "1" in the response. I cannot change type of empType: Option[String] to String (need it that way).
I know it's a very minor change needed but cannot place what.
Please help.
scala playframework
edited May 22 '17 at 9:24
Jasper-M
9,09211329
asked May 22 '17 at 8:57
b1399877b1399877
125
1
For what input do you get"1"as a response?
– Cyrille Corpet
May 22 '17 at 9:25
add a comment |
I am new to scala programming and am facing difficulty to convert from String to Option[String]. I am not sure if the approach I am following is correct or not. Here is the code:
EmpInfo.scala:
package model
case class EmpInfo(empType: Option[String]) extends Named {
override def name: String = clientName
}
EmpReconciliator.scala
import scalaz._
import Scalaz._
import scala.concurrent.Future
import play.api.mvc.Action
import play.api.mvc.AnyContent
import model.EmpInfo
trait EmpReconciliator[A] extends BaseController[A] {
val oldName = "abc"
val newName = "xyz"
def reconcileGeneric(genericResource: A) = {
(genericResource match {
case empInfo : EmpInfo => reconcileEmpInfo(empInfo)
case other@_ => other
}).asInstanceOf[A]
}
def reconcileEmpInfo(empInfo: EmpInfo) = {
empInfo.empTypeType match {
case Some(oldName) => empInfo.copy(empType = Some(newName))
case _ => empInfo
}
}
}
Here, I need to replace all instances of oldName with newName.
But instead, I am getting "1" in the response. I cannot change type of empType: Option[String] to String (need it that way).
I know it's a very minor change needed but cannot place what.
Please help.
scala playframework
edited May 22 '17 at 9:24
Jasper-M
9,09211329
asked May 22 '17 at 8:57
b1399877b1399877
125
I am new to scala programming and am facing difficulty to convert from String to Option[String]. I am not sure if the approach I am following is correct or not. Here is the code:
EmpInfo.scala:
package model
case class EmpInfo(empType: Option[String]) extends Named {
override def name: String = clientName
}
EmpReconciliator.scala
import scalaz._
import Scalaz._
import scala.concurrent.Future
import play.api.mvc.Action
import play.api.mvc.AnyContent
import model.EmpInfo
trait EmpReconciliator[A] extends BaseController[A] {
val oldName = "abc"
val newName = "xyz"
def reconcileGeneric(genericResource: A) = {
(genericResource match {
case empInfo : EmpInfo => reconcileEmpInfo(empInfo)
case other@_ => other
}).asInstanceOf[A]
}
def reconcileEmpInfo(empInfo: EmpInfo) = {
empInfo.empTypeType match {
case Some(oldName) => empInfo.copy(empType = Some(newName))
case _ => empInfo
}
}
}
Here, I need to replace all instances of oldName with newName.
But instead, I am getting "1" in the response. I cannot change type of empType: Option[String] to String (need it that way).
I know it's a very minor change needed but cannot place what.
Please help.
scala playframework
scala playframework
edited May 22 '17 at 9:24
Jasper-M
9,09211329
asked May 22 '17 at 8:57
b1399877b1399877
125
edited May 22 '17 at 9:24
Jasper-M
9,09211329
asked May 22 '17 at 8:57
b1399877b1399877
125
edited May 22 '17 at 9:24
Jasper-M
9,09211329
edited May 22 '17 at 9:24
Jasper-M
9,09211329
edited May 22 '17 at 9:24
Jasper-M
9,09211329
9,09211329
asked May 22 '17 at 8:57
b1399877b1399877
125
asked May 22 '17 at 8:57
b1399877b1399877
125
asked May 22 '17 at 8:57
b1399877b1399877
125
125
1
For what input do you get"1"as a response?
– Cyrille Corpet
May 22 '17 at 9:25
add a comment |
1
For what input do you get"1"as a response?
– Cyrille Corpet
May 22 '17 at 9:25
1
1
For what input do you get
"1" as a response?– Cyrille Corpet
May 22 '17 at 9:25
For what input do you get
"1" as a response?– Cyrille Corpet
May 22 '17 at 9:25
add a comment |
3 Answers
3
active
oldest
votes
I don't understand your question. But if you want to use oldName as a pattern, you have to capitalize it.
val OldName = "abc"
val newName = "xyz"
empInfo.empType match {
case Some(OldName) => empInfo.copy(empType = Some(newName))
case _ => empInfo
}
If that doesn't solve the problem you're having, please state your problem more clearly and precisely.
answered May 22 '17 at 9:30
Jasper-MJasper-M
9,09211329
1
see, also, stackoverflow.com/questions/7078022/…
– Yaneeve
May 22 '17 at 10:24
Type ofSome("a string"):Option[String]. GetStringValue of aOption[String]type:str.get
– Seraf
Aug 24 '17 at 17:29
add a comment |
I think you have to add toString()
here
case Some(oldName) => empInfo.copy(empType =
Some(newName)).toString()
case _ => empInfo
answered May 22 '17 at 9:02
manikandanmanikandan
12
add a comment |
As I understand, the problem that you are facing is to convert a String type to Option[String] type.
Then try wrapping it like this: Option(<variablename>), as in Option(newvalue) in your example.
edited Nov 23 '18 at 17:00
Alonso Dominguez
6,28311936
answered Nov 23 '18 at 12:38
RaviRavi
1
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
I don't understand your question. But if you want to use oldName as a pattern, you have to capitalize it.
val OldName = "abc"
val newName = "xyz"
empInfo.empType match {
case Some(OldName) => empInfo.copy(empType = Some(newName))
case _ => empInfo
}
If that doesn't solve the problem you're having, please state your problem more clearly and precisely.
answered May 22 '17 at 9:30
Jasper-MJasper-M
9,09211329
1
see, also, stackoverflow.com/questions/7078022/…
– Yaneeve
May 22 '17 at 10:24
Type ofSome("a string"):Option[String]. GetStringValue of aOption[String]type:str.get
– Seraf
Aug 24 '17 at 17:29
add a comment |
I don't understand your question. But if you want to use oldName as a pattern, you have to capitalize it.
val OldName = "abc"
val newName = "xyz"
empInfo.empType match {
case Some(OldName) => empInfo.copy(empType = Some(newName))
case _ => empInfo
}
If that doesn't solve the problem you're having, please state your problem more clearly and precisely.
answered May 22 '17 at 9:30
Jasper-MJasper-M
9,09211329
1
see, also, stackoverflow.com/questions/7078022/…
– Yaneeve
May 22 '17 at 10:24
Type ofSome("a string"):Option[String]. GetStringValue of aOption[String]type:str.get
– Seraf
Aug 24 '17 at 17:29
add a comment |
I don't understand your question. But if you want to use oldName as a pattern, you have to capitalize it.
val OldName = "abc"
val newName = "xyz"
empInfo.empType match {
case Some(OldName) => empInfo.copy(empType = Some(newName))
case _ => empInfo
}
If that doesn't solve the problem you're having, please state your problem more clearly and precisely.
answered May 22 '17 at 9:30
Jasper-MJasper-M
9,09211329
I don't understand your question. But if you want to use oldName as a pattern, you have to capitalize it.
val OldName = "abc"
val newName = "xyz"
empInfo.empType match {
case Some(OldName) => empInfo.copy(empType = Some(newName))
case _ => empInfo
}
If that doesn't solve the problem you're having, please state your problem more clearly and precisely.
answered May 22 '17 at 9:30
Jasper-MJasper-M
9,09211329
edited May 22 '17 at 9:35
answered May 22 '17 at 9:30
Jasper-MJasper-M
9,09211329
answered May 22 '17 at 9:30
Jasper-MJasper-M
9,09211329
answered May 22 '17 at 9:30
Jasper-MJasper-M
9,09211329
9,09211329
1
see, also, stackoverflow.com/questions/7078022/…
– Yaneeve
May 22 '17 at 10:24
Type ofSome("a string"):Option[String]. GetStringValue of aOption[String]type:str.get
– Seraf
Aug 24 '17 at 17:29
add a comment |
1
see, also, stackoverflow.com/questions/7078022/…
– Yaneeve
May 22 '17 at 10:24
Type ofSome("a string"):Option[String]. GetStringValue of aOption[String]type:str.get
– Seraf
Aug 24 '17 at 17:29
1
1
see, also, stackoverflow.com/questions/7078022/…
– Yaneeve
May 22 '17 at 10:24
see, also, stackoverflow.com/questions/7078022/…
– Yaneeve
May 22 '17 at 10:24
Type of
Some("a string"): Option[String]. Get String Value of a Option[String] type: str.get– Seraf
Aug 24 '17 at 17:29
Type of
Some("a string"): Option[String]. Get String Value of a Option[String] type: str.get– Seraf
Aug 24 '17 at 17:29
add a comment |
I think you have to add toString()
here
case Some(oldName) => empInfo.copy(empType =
Some(newName)).toString()
case _ => empInfo
answered May 22 '17 at 9:02
manikandanmanikandan
12
add a comment |
I think you have to add toString()
here
case Some(oldName) => empInfo.copy(empType =
Some(newName)).toString()
case _ => empInfo
answered May 22 '17 at 9:02
manikandanmanikandan
12
add a comment |
I think you have to add toString()
here
case Some(oldName) => empInfo.copy(empType =
Some(newName)).toString()
case _ => empInfo
answered May 22 '17 at 9:02
manikandanmanikandan
12
I think you have to add toString()
here
case Some(oldName) => empInfo.copy(empType =
Some(newName)).toString()
case _ => empInfo
answered May 22 '17 at 9:02
manikandanmanikandan
12
answered May 22 '17 at 9:02
manikandanmanikandan
12
answered May 22 '17 at 9:02
manikandanmanikandan
12
answered May 22 '17 at 9:02
manikandanmanikandan
12
12
add a comment |
add a comment |
As I understand, the problem that you are facing is to convert a String type to Option[String] type.
Then try wrapping it like this: Option(<variablename>), as in Option(newvalue) in your example.
edited Nov 23 '18 at 17:00
Alonso Dominguez
6,28311936
answered Nov 23 '18 at 12:38
RaviRavi
1
add a comment |
As I understand, the problem that you are facing is to convert a String type to Option[String] type.
Then try wrapping it like this: Option(<variablename>), as in Option(newvalue) in your example.
edited Nov 23 '18 at 17:00
Alonso Dominguez
6,28311936
answered Nov 23 '18 at 12:38
RaviRavi
1
add a comment |
As I understand, the problem that you are facing is to convert a String type to Option[String] type.
Then try wrapping it like this: Option(<variablename>), as in Option(newvalue) in your example.
edited Nov 23 '18 at 17:00
Alonso Dominguez
6,28311936
answered Nov 23 '18 at 12:38
RaviRavi
1
As I understand, the problem that you are facing is to convert a String type to Option[String] type.
Then try wrapping it like this: Option(<variablename>), as in Option(newvalue) in your example.
edited Nov 23 '18 at 17:00
Alonso Dominguez
6,28311936
answered Nov 23 '18 at 12:38
RaviRavi
1
edited Nov 23 '18 at 17:00
Alonso Dominguez
6,28311936
edited Nov 23 '18 at 17:00
Alonso Dominguez
6,28311936
edited Nov 23 '18 at 17:00
Alonso Dominguez
6,28311936
6,28311936
answered Nov 23 '18 at 12:38
RaviRavi
1
answered Nov 23 '18 at 12:38
RaviRavi
1
answered Nov 23 '18 at 12:38
RaviRavi
1
1
add a comment |
add a comment |
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1
For what input do you get
"1"as a response?– Cyrille Corpet
May 22 '17 at 9:25