Cut only last characters from one file name ((String)globalMap.get("tFileList_1_CURRENT_FILE")

[FSternberg]

Only one doubt, how can I to cut the last characters from one file name using  ((String)globalMap.get("tFileList_1_CURRENT_FILE"))?

Example:

Actual file name:

 DALUNOS_RH

_20211222

Necessary:

DALUNOS_RH

I tried as below but the job show me DALUNOS_RH_202 ....

((String)globalMap.get("tFileList_1_CURRENT_FILE")).substring(0,((String)globalMap.get("tFileList_1_CURRENT_FILE")).length() - 9)

[Anonymous]

You can try expression:

((String)globalMap.get("tFileList_1_CURRENT_FILE")).substring(0, ((String)globalMap.get("tFileList_1_CURRENT_FILE")).lastIndexOf("_"))

[FSternberg]

Hello thanks and only one more doubt, if in case I have as the following: DALUNOS_RH_211203

How can I cut _211203?

thanks a lot

[VenkyM]

1. if the length is fixed width, you can use stringhandling function.
2. if the length is not fixed, you can use delimiter function (delimiter is "_") and take only the first two words.

[Anonymous]

cut the string from last index of "_",

((String)globalMap.get("tFileList_1_CURRENT_FILE")).substring(((String)globalMap.get("tFileList_1_CURRENT_FILE")).lastIndexOf("_"))

[FSternberg1638555481]

Hello,

The lenth is fixed at the end but I need of .CSV for example.

DALUNOS_RH_20211228_125745.CSV

To be: : DALUNOS_RH.CSV

Could you give me one example of the line commnand to keep the .csv?

thanks

[FSternberg]

I solved it thanks