Don't display status code twice in UI errors
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4 changed files with 4 additions and 4 deletions
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@ -30,7 +30,7 @@ const makeRequest = async (method: string, path: string, data?: any): Promise<an
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}
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error.status = response.status;
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error.statusText = response.statusText;
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error.statusText = `${response.status} ${response.statusText}`;
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if (!response.ok) {
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const errorBody = await response.text();
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