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Explicitly return emty link if no 'kind' available

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Signed-off-by: Alex Andreev <alex.andreev.email@gmail.com>
This commit is contained in:
Alex Andreev 2020-08-30 15:44:53 +03:00
parent 1301e6611d
commit 44b74d6a45
1 changed files with 4 additions and 2 deletions
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6
src/renderer/api/kube-api-parse.ts
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@ -117,10 +117,12 @@ export function createKubeApiURL(ref: IKubeApiLinkRef): string {
export function lookupApiLink(ref: IKubeObjectRef, parentObject: KubeObject): string {
const {
kind = "", apiVersion, name,
kind, apiVersion, name,
namespace = parentObject.getNs()
} = ref;
if (!kind) return "";
// search in registered apis by 'kind' & 'apiVersion'
const api = apiManager.getApi(api => api.kind === kind && api.apiVersionWithGroup == apiVersion)
if (api) {
@ -129,7 +131,7 @@ export function lookupApiLink(ref: IKubeObjectRef, parentObject: KubeObject): st
// lookup api by generated resource link
const apiPrefixes = ["/apis", "/api"];
const resource = kind.endsWith("s") ? `${kind}es` : `${kind}s`;
const resource = kind.endsWith("s") ? `${kind.toLowerCase()}es` : `${kind.toLowerCase()}s`;
for (const apiPrefix of apiPrefixes) {
const apiLink = createKubeApiURL({ apiPrefix, apiVersion, name, namespace, resource });
if (apiManager.getApi(apiLink)) {