Par*_*eri 8 dart firebase flutter google-cloud-functions
不知道如何从颤振中的云函数中获得响应。
我的云功能
const functions = require('firebase-functions');
const admin = require('firebase-admin');
admin.initializeApp();
exports.testDemo = functions.https.onRequest((request, response) => {
return response.status(200).json({msg:"Hello from Firebase!"});
});
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我的颤振代码
///Getting an instance of the callable function:
try {
final HttpsCallable callable = CloudFunctions.instance.getHttpsCallable(
functionName: 'testDemo',);
///Calling the function with parameters:
dynamic resp = await callable.call();
print("this is responce from firebase $resp");
} on CloudFunctionsException catch (e) {
print('caught firebase functions exception');
print(e.code);
print(e.message);
print(e.details);
} catch (e) {
print('caught generic exception');
print(e);
}
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颤振:捕获了 Firebase 函数异常颤振:内部颤振:响应缺少数据字段。颤振:空
用
exports.testDemo = functions.https.onCall((data, context) => {
return {msg:"Hello from Firebase!"};
});
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调用函数时需要添加参数:
改变:
// Calling a function without parameters is a different function!
dynamic resp = await callable.call();
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到:
dynamic resp = await callable.call(
<String, dynamic>{
'YOUR_PARAMETER_NAME': 'YOUR_PARAMETER_VALUE',
},
);
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描述在这里
然后打印响应:
print(resp.data)
print(resp.data['msg'])
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此处和此处的 Firebase Functions for flutter 示例
您必须明确地将属性“data”放入响应的 json 中。
喜欢:
response.send({
"status" : success,
"data" : "some... data"
});
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