此问题已删除
答案 0 :(得分:1)
您需要该属性,然后可以返回检查结果。
function isOver40(user) {
return user.age > 40;
}
使用What Methods are considered Keywords:
function isOver40({ age }) {
return age > 40;
}
答案 1 :(得分:0)
传递对象时,必须访问对象的属性,而不是对象本身。说你有:
UPDATE tmp_elliottb.UpdateTable
SET my_field = ARRAY(
SELECT AS STRUCT * REPLACE (IF(false, struct_to_set_null, NULL) AS struct_to_set_null )
FROM UNNEST(my_field)
)
WHERE true;
您的函数应如下图所示
user { //javascript object
name: "foobar",
age: 25
}
函数调用:
function isOver40 (user) {
if (user.age > 40) {return true}
else {return false} //you can just say return false here, as the else is not really needed
}
希望这会有所帮助