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Resolves #459 -- parent ordering can break some operations

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Peter Cottle 2018-04-09 08:34:01 -07:00
parent 69b99b82d0
commit b6a34ec17e
3 changed files with 19 additions and 10 deletions
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27
src/js/git/index.js
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@ -418,12 +418,24 @@ GitEngine.prototype.makeRemoteBranchForRemote = function(branchName) {
};
GitEngine.prototype.findCommonAncestorForRemote = function(myTarget) {
// like the method below but opposite
while (!this.origin.refs[myTarget]) {
var parents = this.refs[myTarget].get('parents');
myTarget = parents[0].get('id');
if (this.origin.refs[myTarget]) {
return myTarget;
}
return myTarget;
var parents = this.refs[myTarget].get('parents');
if (parents.length === 1) {
// Easy, we only have one parent. lets just go upwards
myTarget = parents[0].get('id');
// Recurse upwards to find where our remote has a commit.
return this.findCommonAncestorForRemote(myTarget);
}
// We have multiple parents so find out where these two meet.
var leftTarget = this.findCommonAncestorForRemote(parents[0].get('id'));
var rightTarget = this.findCommonAncestorForRemote(parents[1].get('id'));
return this.getCommonAncestor(
leftTarget,
rightTarget,
true // dont throw since we dont know the order here.
).get('id');
};
GitEngine.prototype.findCommonAncestorWithRemote = function(originTarget) {
@ -2100,7 +2112,6 @@ GitEngine.prototype.hgRebase = function(destination, base) {
GitEngine.prototype.rebase = function(targetSource, currentLocation, options) {
// first some conditions
debugger;
if (this.isUpstreamOf(targetSource, currentLocation)) {
this.command.setResult(intl.str('git-result-uptodate'));
@ -2764,8 +2775,8 @@ GitEngine.prototype.log = function(ref, omitSet) {
});
};
GitEngine.prototype.getCommonAncestor = function(ancestor, cousin) {
if (this.isUpstreamOf(cousin, ancestor)) {
GitEngine.prototype.getCommonAncestor = function(ancestor, cousin, dontThrow) {
if (this.isUpstreamOf(cousin, ancestor) && !dontThrow) {
throw new Error('Dont use common ancestor if we are upstream!');
}