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Simplify provider display name logic and add debug logging

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- Replace complex fallback chain for provider display name with direct provider name access
- Add console.log for model debugging in getModelName function
This commit is contained in:
suyao 2025-09-20 16:23:37 +08:00
parent 117e390cf1
commit 282aa6e81a
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2 changed files with 2 additions and 8 deletions
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8
src/main/apiServer/utils/index.ts
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@ -190,13 +190,7 @@ export async function validateModelId(
export function transformModelToOpenAI(model: Model, providers: Provider[]): ApiModel { export function transformModelToOpenAI(model: Model, providers: Provider[]): ApiModel {
const provider = providers.find((p) => p.id === model.provider) const provider = providers.find((p) => p.id === model.provider)
const providerDisplayName = const providerDisplayName = provider?.name
provider?.name ??
(provider as { providerName?: string })?.providerName ??
(provider as { displayName?: string })?.displayName ??
(model as { providerName?: string; providerDisplayName?: string })?.providerName ??
(model as { providerDisplayName?: string })?.providerDisplayName ??
provider?.id
return { return {
id: `${model.provider}:${model.id}`, id: `${model.provider}:${model.id}`,
object: 'model', object: 'model',

2
src/renderer/src/services/ModelService.ts
View File

@ -21,7 +21,7 @@ export const hasModel = (m?: Model) => {
export function getModelName(model?: Model | ApiModel) { export function getModelName(model?: Model | ApiModel) {
const modelName = model?.name || model?.id || '' const modelName = model?.name || model?.id || ''
console.log(model)
// For API models that have provider_name field, use it directly // For API models that have provider_name field, use it directly
const apiModel = model as ApiModel const apiModel = model as ApiModel
if (apiModel?.provider_name) { if (apiModel?.provider_name) {