Question3Replies73Views LAKSHMIM7752 Member since 2019 1 post S & P Posted: November 12, 2019Last activity: November 14, 2019 Closed Replace only Name with name not LastName with Lastname in JSON Stringin my JSON String i want to Name with name but when i am using replaceAll function it is replacing LastName with Lastname i dont wnat that. Can you please help me to resolve this issue.***Moderator Edit-Vidyaranjan: Updated Platform Capability*** Data Integration User Interface ×Close popoverFacebookTwitterLinkedinEmail Copy Link Copied! Moderation Team has archived post This thread is closed to future replies. Content and links will no longer be updated. If you have the same/similar Question, please write a new Question. Posted: 1 year agosahuv1 PEGA replied to LAKSHMIM7752Hi , Please share the details of this usecase , i.e where you have tried this? Are you creating your own JSON and trying it some where ? Version of the pega will also help us. Posted: 1 year agovaspoz BPM Company replied to LAKSHMIM7752Possible solutions: 1. Replase not "Name" to "name", but " Name" to " name" with a space in the beginning 2. Firstly, replase all "LastName" to something like "LastNam", then do your replase all "Name" to "name". At the end replace back "LastNam" to "LastName" Posted: 1 year agobagas PEGA replied to LAKSHMIM7752If you want to do this in integration layer, you can try with pzExternalName property qualifier for the property Name to map it as "name" while generating JSON or the other way. If you want to do pure string manipulation, then you can try different options (like what Vaspoz suggested here).