java.lang.NumberFormatException when trying to fetch a line in a Cassandra table

Question

I use the Datastax driver to fetch a line in a Cassandra table :

 MappingManager manager = new MappingManager(session);
 Mapper<CassandraEntity> mapper = manager.mapper(CassandraEntity.class);
 UUID id = UUID.fromString(uuid);    
 CassandraEntity cassandraEntity = mapper.get(id);

When the uuid exists in the table then everything works fine. But when the uuid does not exist I have this error :

ERROR [2017-01-27 14:27:24,030] io.dropwizard.jersey.errors.LoggingExceptionMapper: Error handling a request: 48a3bf442525acf8
! java.lang.NumberFormatException: For input string: "68c34e83db3o"
! at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65) ~[na:1.7.0_75]
! at java.lang.Long.parseLong(Long.java:441) ~[na:1.7.0_75]
! at java.lang.Long.valueOf(Long.java:513) ~[na:1.7.0_75]
! at java.lang.Long.decode(Long.java:665) ~[na:1.7.0_75]
! at java.util.UUID.fromString(UUID.java:206) ~[na:1.7.0_75]

How can I nicely manage this error ?

score 2 · Answer 1 · edited Jun 20 '20 at 09:12

2

68c34e83db3o is not a valid UUID so UUID.fromString is throwing this exception, the javadocs state:

Throws:

IllegalArgumentException - If name does not conform to the string ? representation as described in toString()

NumberFormatException is an instance of IllegalArgumentExcepiton.

How are you forming uuid before passing it into fromString? It's possible you might need to do some formatting/santization before calling fromString. 68c34e83db3o almost looks like a hex string (except for that last o).

edited Jun 20 '20 at 09:12

Community

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answered Jan 27 '17 at 18:03

Andy Tolbert

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Thank you for this answser. I use Dropwizard to invoke the fetch in database. I'm not forming the uuid before passing it into fromString. The uuid is the parameter of the method connect(@QueryParam("uuid") String uuid). The UUID c37d661d-7e61-49ea-96a5-68c34e83db3a exists in my table. So when I hit localhost:8080/connect?uuid=c37d661d-7e61-49ea-96a5-68c34e83‌‌db3a I have no error. On the contrary the UUID c37d661d-7e61-49ea-96a5-68c34e83db3o does not exist in my table. Then when I hit localhost:8080/connect?uuid=c37d661d-7e61-49ea-96a5-68c34e83‌‌db3o I have the error I have described – Denis Jan 30 '17 at 09:13

score 1 · Accepted Answer · edited May 23 '17 at 10:29

c37d661d-7e61-49ea-96a5-68c34e83‌‌‌db3o is not a valid UUID. The sixteen octets of a UUID are represented as 32 lowercase hexadecimal (base 16) digits, displayed in five groups separated by hyphens, in the form 8-4-4-4-12 for a total of 36 characters (32 alphanumeric characters and four hyphens). So :

c37d661d-7e61-49ea-96a5-68c34e83‌‌‌db3a is a valid UUID
c37d661d-7e61-49ea-96a5-68c34e83‌‌‌db3b is a valid UUID
c37d661d-7e61-49ea-96a5-68c34e83‌‌‌db3c is a valid UUID
c37d661d-7e61-49ea-96a5-68c34e83‌‌‌db3d is a valid UUID
c37d661d-7e61-49ea-96a5-68c34e83‌‌‌db3e is a valid UUID
c37d661d-7e61-49ea-96a5-68c34e83‌‌‌db3f is a valid UUID

But c37d661d-7e61-49ea-96a5-68c34e83‌‌‌db3g is not a valid UUID nor c37d661d-7e61-49ea-96a5-68c34e83‌‌‌db3o

To quickly check if a UUID is valid you can use this Fiddle : https://jsfiddle.net/mshaffer/6e2Ljh97

As Andy Tolbert said, when the UUID is not valid then the UUID.fromString method throws an IllegalArgumentException. To handle this exception, use the solution given at How to judge a string is UUID type? :

try{
    UUID uuid = UUID.fromString(someUUID);
    //do something
} catch (IllegalArgumentException exception){
   //handle the case where string is not valid UUID 
}

java.lang.NumberFormatException when trying to fetch a line in a Cassandra table

2 Answers2