在乌鸦中获取非Int(字符串)URL资源ID
Fetch non-int (string) url resource-id in crow
如何在C 乌鸦中获得非INT Resource-ID。我无法添加路线 CROW_ROUTE(app, "/uid/<std::string>")
或CROW_ROUTE(app, "/uid/<char*>")
无法编译。Donot有这样的案例。我尝试了
int main() {
crow::SimpleApp app;
CROW_ROUTE(app, "/uid/*").methods("GET"_method)
([](const crow::request& req){
return "hello";
});
CROW_ROUTE(app, "/uid/<int>").methods("GET"_method)
([](const crow::request& req, int id){
return std::to_string(id);
});
app.port(8888).run();
}
,但它们都没有(尽管正确地(拦截GET /uid/uid_123 HTTP/1.1
(资源为字符串"uid_123"
(
对于以下Python代码,我想在C 乌鸦库中获得它
from flask import Flask
app = Flask(__name__)
@app.route("/uid/<path:path>")
def hello1(path):
print ("it is path is ", path)
return "user id is -" + path
if __name__ == "__main__":
app.run(host="0.0.0.0", port=8888)
上面有任何解决方法吗?
最初由erow和jihoo在github上回答
CROW_ROUTE(app,"/uid/<path>")
([](std::string path){
return path;
});
除了int
和path
之外,也有效:uint
,double
和string