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Use solo la variable que provocó que se activara una declaración if, dentro de la declaración if

digamos que tiene una sentencia if:

 try { if (!a || !b || !c || !d) { let nullVariable = ???; // use the variable that is null throw nullVariable } } catch (ex) { log.debug(`${ex} is not defined`); }

¿Existe una forma integrada de ver qué variable se estableció en nulo, sin crear una declaración if para cada variable individual?

about 4 years ago · Juan Pablo Isaza
2 answers
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0

Puede poner la asignación dentro de la expresión de condición if .

 let a, b, c, d; a = 3; b = "foo"; d = {x: 10}; let nullVariable = (!a && 'a') || (!b && 'b') || (!c && 'c') || (!d && 'd'); if (nullVariable) { console.log(`${nullVariable} is not defined`); }

about 4 years ago · Juan Pablo Isaza Report

0

Puede extraer el nombre de la variable usando la abreviatura de objeto.

 const fn = ({ a, b, c, d }) => { let nullVariable = Object.entries({ a, b, c, d }).find(([_, val]) => !val)?.[0]; if(nullVariable) console.log(nullVariable + ' is falsy'); }; fn({ a: 1, c: 2, d: 3 }); fn({ a: 1, b: 2, c: 3 }); fn({ a: 1, b: 2, c: 3, d: 4 });

about 4 years ago · Juan Pablo Isaza Report
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