Only leave form if API request succeeded

If the "Ask to join" request failed for any reason, we don't want to navigate away from the form.
sillsdev · Aug 5, 2024 · 6539a7a · 6539a7a
1 parent 66d11a4
commit 6539a7a
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Showing 2 changed files with 6 additions and 5 deletions.
diff --git a/frontend/src/routes/(authenticated)/project/create/+page.svelte b/frontend/src/routes/(authenticated)/project/create/+page.svelte
@@ -172,10 +172,12 @@
   let showRelatedProjects = true;
 
   async function askToJoinProject(projectId: string, projectName: string): Promise<void> {
-    await _askToJoinProject(projectId);
-    notifySuccess($t('project.create.join_request_sent', { projectName }))
-    $tainted = undefined; // Prevent "are you sure you want to leave?" warning
-    await goto('/');
+    const joinResult = await _askToJoinProject(projectId);
+    if (!joinResult.error) {
+      notifySuccess($t('project.create.join_request_sent', { projectName }))
+      $tainted = undefined; // Prevent "are you sure you want to leave?" warning
+      await goto('/');
+    }
   }
 </script>
 

diff --git a/frontend/src/routes/(authenticated)/project/create/+page.ts b/frontend/src/routes/(authenticated)/project/create/+page.ts
@@ -127,5 +127,4 @@ export async function _askToJoinProject(projectId: string): $OpResult<AskToJoinP
     `),
     { input: { projectId } });
   return result;
-  //
 }
-Original file line number
+Diff line change
@@ Expand Up @@
         `),
         { input: { projectId } });
       return result;
-      //
     }