Non*_*one 0 javascript typescript
我有这个对象:
this.prepaidObject = {
'customerType' : this.prepaidDetailForm.prepaidDetails.customerType,
'firstName' : this.prepaidDetailForm.prepaidDetails.firstName,
'lastName' : this.prepaidDetailForm.prepaidDetails.lastName,
'note' : this.prepaidDetailForm.prepaidDetails.note,
'created': this.prepaidDetailForm.prepaidDetails.created
};
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现在有时某些属性是未定义的。我想要的是如果this.prepaidDetailForm.prepaidDetails未定义属性之一不显示它们。因此,例如,如果this.prepaidDetailForm.prepaidDetails.firsName未定义,我不需要创建'firstName'属性 isnide 对象。任何建议我该怎么做?
检查您的对象:
for( var m in this.prepaidObject ) {
if ( this.prepaidObject[m] == undefined ) {
delete this.prepaidObject[m];
}
}
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执行此操作的最短方法是在parse后进行stringify。
确切的语法:
let prepaidObjectNew = JSON.parse(JSON.stringify(prepaidObject))
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例子
前:
var prepaidObject = {
'customerType' : 'TestCustType',
'firstName' : "sample FirstName",
'lastName' : "sample LastName",
'note' : undefined
};
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后:
{
'customerType' : 'TestCustType',
'firstName' : "sample FirstName",
'lastName' : "sample LastName"
};
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