Cypress.io – Programmatically set response based on request parameters in cy.route()
I am stubbing an api request in my end-to-end test and would like to be able to return a different response depending on the request parameters that are sent out.
Currently, my stub is returning a static response regardless of what is posted, and looks like this:
cy.server()
cy.route({
method: 'POST',
url: '**/redeem-code',
status: 200,
response: {
status: "Success"
},
delay: 500
})
I would like to be able to check the posted request parameters and then conditionally decide which response to return. I’m trying to do something like this:
cy.server()
cy.route({
method: 'POST',
url: '**/redeem-code',
status: 200,
response: (req) => {
if(req.code == '1234') return { status: "Success" }
else return { status: "Failure" }
},
delay: 500
})
Obviously, the code above doesn’t work; it’s just an example of what I’m trying to do.
I know Cypress allows for response methods, but I can’t find the syntax for what I’m wanting to do anywhere in their docs. How do I get the request parameters in my response method so I can decide which response to return?
This is sadly currently not supported with cy.server.
The issue is being tracked here : https://github.com/cypress-io/cypress/issues/521
Workaround
Use standard javascript mocking. You can run this mocks in tests by using cypress onBeforeLoad, mentioned a few times in the linked issue. Its not pretty. Hopefully cypress gets native support in cy.server.
I had the same problem and made a feature to enable this. It does require a bounce back url to be stood up but the code install instructions can be seen here https://bitbucket.org/snippets/matt-tasc/onraxo
I believe this should work
cy.server({
onResponse: ({ status, url, response }) => {
if(url !== 'yoururl') return response;
return {
...response,
body: { status: status === 1234 ? 'success' : 'failure' }
};
}
});
