Sign Up

Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.

Have an account? Sign In

Have an account? Sign In Now

Sign In

Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.

Sign Up Here

Forgot Password?

Don't have account, Sign Up Here

Forgot Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

Have an account? Sign In Now

You must login to ask a question.

Forgot Password?

Need An Account, Sign Up Here

Please briefly explain why you feel this question should be reported.

Please briefly explain why you feel this answer should be reported.

Please briefly explain why you feel this user should be reported.

Sign InSign Up

The Archive Base

The Archive Base Logo The Archive Base Logo

The Archive Base Navigation

  • SEARCH
  • Home
  • About Us
  • Blog
  • Contact Us
Search
Ask A Question

Mobile menu

Close
Ask a Question
  • Home
  • Add group
  • Groups page
  • Feed
  • User Profile
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Buy Points
  • Users
  • Help
  • Buy Theme
  • SEARCH
Home/ Questions/Q 8447967
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: June 10, 20262026-06-10T10:18:18+00:00 2026-06-10T10:18:18+00:00

Abstract: To reproduce the error create a branch and check it out let someone

  • 0

Abstract: To reproduce the error

  • create a branch and check it out
  • let someone else delete it and create a new branch with the same name
  • now do git branch -D <branch> and git checkout -b <branch> --track origin/<branch>
  • on a git pull you get ! [rejected] <branch> -> origin/<branch> (non-fast-forward)

to fix it, you have to delete the remote tracking information with git branch -d -r origin/<branch> as well


OLD: Someone deleted the develop branch and created it to remove all feature branches and have the master as base again. Then he added some of the feature branches but not some others that made problems.

I did a git branch -D develop and git checkout -b develop --track origin/develop.

When i now try git pull i get a ! [rejected] develop -> origin/develop (non-fast-forward)

a git remote show origin shows

Local refs configured for 'git push': 
develop       pushes to develop     (local out of date)

i can now do a git fetch origin develop and git merge FETCH_HEAD but then i have some conflicts and he wants to push a lot of things to develop. (maybe the old branch commits?) And with a git reset --hard i’m back where the git pull shows the rejected message ..

How do i checkout the recreated branch best?

EDIT: even when i do git branch -D develop i get with git pull ! [rejected] develop -> origin/develop (non-fast-forward) and git remote show origin said everything (up to date)

EDIT: i didn’t recognized it at first, because the commit message was the same, but after a reset the HEAD is on a sha that the remote does not have, so still on the “old” branch ?

  • 1 1 Answer
  • 0 Views
  • 0 Followers
  • 0
Share
  • Facebook
  • Report

Leave an answer
Cancel reply

You must login to add an answer.

Forgot Password?

Need An Account, Sign Up Here

1 Answer

  • Voted
  • Oldest
  • Recent
  • Random
  1. Editorial Team
    Editorial Team
    2026-06-10T10:18:19+00:00Added an answer on June 10, 2026 at 10:18 am

    i had to delete the tracking branch as well

    git branch -d -r origin/develop
    
    • 0
    • Reply
    • Share
      Share
      • Share on Facebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
      • Report

Sidebar

Related Questions

public abstract class Master { public void printForAllMethodsInSubClass() { System.out.println (Printing before subclass method
Abstract problem: Define some (non-rectangular, non-circular) topologically closed area on a map. Find out
abstract class db_table { static function get_all_rows() { ... while(...) { $rows[] = new
abstract class SettingSaver { public abstract void add(string Name, string Value); public abstract void
abstract class A{ abstract test(); function __construct (){ //check if test method exists on
abstract class A { public void methodA() { System.out.println(methodA); methodB(); showName(); } public abstract
public abstract class Animal , IValidatableObject { public string Id {get;set;} public string Name
abstract class MyClass{ static class StaticClass{ public void showAdd(){ System.out.println( show add); } }
Check below for UPDATE, I could reproduce and need help. I have a strange
abstract method foo(E) in wtf.is.going.on cannot be accessed directly super.foo(e) I get that error

Explore

  • Home
  • Add group
  • Groups page
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Users
  • Help
  • SEARCH

Footer

© 2021 The Archive Base. All Rights Reserved
With Love by The Archive Base

Insert/edit link

Enter the destination URL

Or link to existing content

    No search term specified. Showing recent items. Search or use up and down arrow keys to select an item.