Fix package name handling to retain version and strip ‘@’ suffix

Signed-off-by: Emin Aktas <[email protected]>

Author: eminaktas

pkg/build/lock.go

@@ -134,9 +134,10 @@ func unify(originals []string, inputs []resolved) (map[string][]string, map[stri
 		// special characters that delimit the package name from the cosntraint
 		// so lop off the package name and stick the rest of the constraint into
 		// the versions map.
-		if idx := strings.IndexAny(orig, "=<>~"); idx >= 0 {
+		if idx := strings.IndexAny(orig, "@=<>~"); idx >= 0 {
 			name = orig[:idx]
 		}
+
 		originalPackages.packages.Insert(name)
 		originalPackages.versions[name] = strings.TrimPrefix(orig, name)
 	}
@@ -235,7 +236,12 @@ func unify(originals []string, inputs []resolved) (map[string][]string, map[stri
 	// Append all of the resolved and unified packages with an exact match
 	// based on the resolved version we found.
 	for _, pkg := range sets.List(acc.packages) {
-		pl = append(pl, fmt.Sprintf("%s=%s", pkg, acc.versions[pkg]))
+		pkgName := fmt.Sprintf("%s=%s", pkg, acc.versions[pkg])
+		// Add back the version if that is stripped before
+		if originalPackages.versions[pkg] != "" {
+			pkgName = pkgName + originalPackages.versions[pkg]
+		}
+		pl = append(pl, pkgName)
 	}
 	// Sort the package list explicitly with the `=` included.
 	// This is because (foo, foo-bar) sorts differently than (foo=1, foo-bar=1)
@@ -249,7 +255,12 @@ func unify(originals []string, inputs []resolved) (map[string][]string, map[stri
 	for _, input := range inputs {
 		pl := make([]string, 0, len(input.packages))
 		for _, pkg := range sets.List(input.packages) {
-			pl = append(pl, fmt.Sprintf("%s=%s", pkg, input.versions[pkg]))
+			pkgName := fmt.Sprintf("%s=%s", pkg, input.versions[pkg])
+			// Add back the version if that is stripped before
+			if originalPackages.versions[pkg] != "" {
+				pkgName = pkgName + originalPackages.versions[pkg]
+			}
+			pl = append(pl, pkgName)
 		}
 		// Sort the package list explicitly with the `=` included.
 		// This is because (foo, foo-bar) sorts differently than (foo=1, foo-bar=1)