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¿Es posible hacer que xhr.onerror se dispare implícitamente?

Estoy creando un ayudante para cargar archivos en React Native, aunque no puedo disparar xhr.onerror, esperaba arrojar un error dentro de onreadystatechange al verificar el estado y el onerror se activaría, pero desafortunadamente no es así. Cualquier buen enfoque de cómo puedo manejar esto?

Aquí está el código completo:

 sendXHR = async ( url: string, formData: FormData, eventListeners: XMLHttpRequestEventTarget, ) => { let xhr = new XMLHttpRequest(); xhr.open('POST', `${API_URL}${url}`, true); const credentials: any = await retrieveToken(); xhr.setRequestHeader('Authorization', `Bearer ${credentials.password}`); xhr.upload.onabort = eventListeners.onabort; xhr.upload.onprogress = eventListeners.onprogress; xhr.upload.onerror = eventListeners.onerror; xhr.upload.onloadstart = eventListeners.onloadstart; xhr.onreadystatechange = () => { if (xhr.readyState === XMLHttpRequest.DONE) { const status = xhr.status; if (status < 200 || status >= 300) { throw new Error(xhr.statusText); } } }; xhr.send(formData); return { response: new Promise((resolve, reject) => { xhr.onload = () => { if (xhr.status >= 200 && xhr.status < 300) { resolve(xhr.response); } else { reject({status: xhr.status, message: xhr.statusText}); } }; xhr.onerror = () => { reject({status: xhr.status, message: xhr.statusText}); }; }), xhr, }; };
about 3 years ago · Juan Pablo Isaza
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