Question 3 Replies 81 Views × Close popover LAKSHMI MNSJ (LAKSHMIM7752) S & P S & P IN View Profile Send Message LAKSHMIM7752 Member since 2019 2 posts S & P Posted: November 12, 2019 Last activity: November 14, 2019 Posted: 12 Nov 2019 13:29 EST Last activity: 14 Nov 2019 0:30 EST Closed Replace only Name with name not LastName with Lastname in JSON String in 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 Experience × Close popover Facebook Twitter Linkedin Email Copy Link Copied! Moderation Team has archived post, learn more × Close popover 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 ago Posted: 12 Nov 2019 23:01 EST × Close popover Vikash Kumar Sahu (sahuv1) PEGA Principal Technical Solution Engineer - India Pegasystems Inc. IN View Profile Send Message sahuv1 PEGA replied to LAKSHMIM7752 Hi , 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 ago Posted: 13 Nov 2019 9:39 EST × Close popover Basil (Vasilii) Pozdeev (vaspoz) BPM Company Senior System Architect BPM Company NL View Profile Send Message vaspoz BPM Company replied to LAKSHMIM7752 Possible 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 ago Posted: 14 Nov 2019 0:30 EST × Close popover Santhosh Bagannagari (bagas) PEGA Tech Lead, Security Engineering Pegasystems Inc. IN View Profile Send Message bagas PEGA replied to LAKSHMIM7752 If 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).