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Arreglos transversales por forEach

Intentando atravesar una matriz en un objeto con clave/par (sí, sé de reducir):

Esta es la matriz:

 filtered2 = [["deviceName", "TestDevice00003"], ["hwModelName", "TestHwModel03"], ["deviceTypeName", "TestDeviceType03"], ["serviceTag", "A1A03"]]

Pero por alguna razón cuando hago forEach:

 filtered2.forEach( (indiv) => { console.log([indiv[0]]+" | " +[indiv[1]]) obj3 = { ...obj3, [[indiv[0]][0]] : [indiv[1]] } })

Console.log verá cada (indiv) como "hwModelName | TestHwModel03" en [indiv[0]]+" | " +[indiv[1]] , por lo que ambas son cadenas básicas

Pero la función forEach ve [indiv[1]] como Array ["TestHwModel03"] Así que tengo que ir un nivel de profundidad más [indiv[1]][0] en la matriz para obtener la cadena sin envolver "TestHwModel03"

¿Por qué existe esta diferencia?

about 4 years ago · Juan Pablo Isaza
2 answers
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0

Quite los corchetes que rodean el valor.

 filtered2 = [["deviceName", "TestDevice00003"], ["hwModelName", "TestHwModel03"], ["deviceTypeName", "TestDeviceType03"], ["serviceTag", "A1A03"]] let obj3 = {} filtered2.forEach( (indiv) => { obj3 = { ...obj3, [indiv[0]] : indiv[1] } }) console.log(obj3)

Aunque es mucho más simple usar Object.fromEntries :

 filtered2 = [["deviceName", "TestDevice00003"], ["hwModelName", "TestHwModel03"], ["deviceTypeName", "TestDeviceType03"], ["serviceTag", "A1A03"]] let obj3 = Object.fromEntries(filtered2) console.log(obj3)

about 4 years ago · Juan Pablo Isaza Report

0

Puede usar el método de reducción en una matriz para convertir la matriz en un mapa. O puede usar Object.fromEntries si su entorno es compatible con la última versión de ES2017.

 filtered2 = [ ["deviceName", "TestDevice00003"], ["hwModelName", "TestHwModel03"], ["deviceTypeName", "TestDeviceType03"], ["serviceTag", "A1A03"], ]; const toMap = (arr = []) => arr.reduce((acc, [key, value]) => { acc[key] = value; return acc; }, {}); console.log(toMap(filtered2)) /// Other way console.log(Object.fromEntries(filtered2))

about 4 years ago · Juan Pablo Isaza Report
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