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Cómo esperar a que todos los procesos terminen en forEach antes de devolver el resultado (Node.js)

Quiero que la función forEach se complete antes de devolver los resultados, ¿cómo se hace esto? La función devuelve nulo en todos los resultados cuando devuelvo el código porque no espera a que finalice cada uno, entonces, ¿cómo resuelvo esto?

 async function fileResult(fileCard, userId, req, fileTypes) { if (req.files && Object.keys(req.files).length) { let resultObj = { smallResult: null, mediumResult: null, largeResult: null, thumbnailResult: null, }; const result = fileTypes.forEach(async (fileType) => { if (req.files[fileType]) { try { const fileUploadResult = await uploadFile(fileCard, userId, req, fileType); if (fileType === 'small') { resultObj['smallResult'] = fileUploadResult; } else if (fileType === 'medium') { resultObj['mediumResult'] = fileUploadResult; } else if (fileType === 'large') { resultObj['largeResult'] = fileUploadResult; } else if (fileType === 'thumbnail') { resultObj['thumbnailResult'] = fileUploadResult; } if (!fileUploadResult.success) { console.log(fileUploadResult.error, 500); return; } // console.log(resultObj); } catch (err) { console.log(err); } } }); return resultObj; } }
about 4 years ago · Juan Pablo Isaza
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Probar esto

 async function fileResult(fileCard, userId, req, fileTypes) { if (req.files && Object.keys(req.files).length) { let resultObj = { smallResult: null, mediumResult: null, largeResult: null, thumbnailResult: null, }; await new Promise((resolve, reject) => { const result = fileTypes.forEach(async (fileType) => { if (req.files[fileType]) { try { const fileUploadResult = await uploadFile(fileCard, userId, req, fileType); if (fileType === 'small') { resultObj['smallResult'] = fileUploadResult; } else if (fileType === 'medium') { resultObj['mediumResult'] = fileUploadResult; } else if (fileType === 'large') { resultObj['largeResult'] = fileUploadResult; } else if (fileType === 'thumbnail') { resultObj['thumbnailResult'] = fileUploadResult; } if (!fileUploadResult.success) { console.log(fileUploadResult.error, 500); resolve("done") return; } // console.log(resultObj); } catch (err) { console.log(err); reject(err) } } }); }) return resultObj; } }

Por favor, acéptelo como respuesta si le funciona, ¡gracias!

about 4 years ago · Juan Pablo Isaza Report
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