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ptaylor
5 years agoEmployee
Here’s a simpler expression that will work with any number of keys without modification:
$array1.map(e => e.entries().sort().toObject(x => x[0], x => x[1]))
Here’s a simpler expression that will work with any number of keys without modification:
$array1.map(e => e.entries().sort().toObject(x => x[0], x => x[1]))