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Javascript: cómo eliminar objetos dentro de un objeto de matrices de objetos por valor clave

Estoy tratando de descubrir cómo eliminar objetos dentro de una matriz en función de un valor clave.

Actualmente, el objeto se parece a esto:

 const myArray = [ { firstname: 'John', lastName: 'Doe', sortOrder: 22 }, { firstname: 'Jane', lastName: 'Doe', sortOrder: 11 },{ firstname: 'Jen', lastName: 'SomeLastName', sortOrder: 33 }, { firstname: 'ExampleFirstName', lastName: 'ExampleLastName', sortOrder: 12 } ];

Espero reducir la matriz en una cantidad específica, eliminando objetos con el orden de clasificación más alto, para que la función de eliminación de objetos regrese:

 const myArray = [ { firstname: 'Jane', lastName: 'Doe', sortOrder: 11 },{ firstname: 'ExampleFirstName', lastName: 'ExampleLastName', sortOrder: 12 } ];

Mi función actual se ve así, pero no elimina nada:

 reduceArray = (myArray, amount) => { let keys = Object.keys(myArray); let keyLength = keys.length; let orderArray = []; let i = 0, j = 0; for (i = 0; i < keyLength; i++) { let subArray = myArray[keys[i]]; let subLength = subArray.length; for (j = 0; j < subLength; j++) { orderArray.push(subArray[j].sortOrder); } } let totalLength = orderArray.length; for (i = 0; i < totalLength - 1; i++) for (j = i + 1; j < totalLength; j++) if (orderArray[i] < orderArray[j]) { let temp = orderArray[i]; orderArray[i] = orderArray[j]; orderArray[j] = temp; } orderArray.splice(amount, totalLength - amount); let deletedAmount = 0; for (i = 0; i < keyLength; i++) { let subArray = myArray[keys[i]]; let subLength = subArray.length; for (j = 0; j < subLength; j++) { if (orderArray.includes(subArray[j].sortOrder)) { subArray.splice(j, 1); j--; subLength--; deletedAmount++; if (deletedAmount == amount) break; } } if (deletedAmount == amount) break; } return myArray; }

Y lo estoy llamando así:

 const reducedArray = reduceArray(myArray, 2);

Simplemente devuelvo mi misma matriz sin quitar nada.

¿Alguna idea de lo que podría estar haciendo mal?

about 4 years ago · Juan Pablo Isaza
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