3

我已经坚持了一段时间。我需要测试一个网站，我需要发布信息以测试它是否出现在页面上。

到目前为止我所拥有的是这个

(async () => {
    const browser = await webkit.launch();
    const page = await browser.newPage();
    await page.route('http://100.100.100.100/', route => route.fulfill({
        status: 200,
        body: body,
    }));
    await page.goto('https://theurlofmywebsite/');
    await page.click('button')
    await page.click('text=Login with LoadTest')
    await page.fill('#Username','username')
    await page.fill('#Password','password')
    await page.click('#loginButton')
    // await page.waitForSelector('text=Dropdown');
    await page.click('css=span >> text=Test')
    await page.click('#root > div > div > header > ul.nav.navbar-nav.area-tabs > li:nth-child(6) > a','Test')
    await page.waitForSelector('text=Detail')
    await page.screenshot({ path: `example3.png` })
    await browser.close();
})();

const body = [ my json post request ]
4

2 回答 2

3 
jest.setTimeout(1000000);
let browser: any;
let page: any;
beforeAll(async () => {
    browser = await chromium.launch();
});
afterAll(async () => {
    await browser.close();
});
beforeEach(async () => {
    page = await browser.newPage();
});
afterEach(async () => {
    await page.close();
});



it("should work", async () => {
    await fetch("http://YOUAWESOMEURL", {
        method: "post",
        body: JSON.stringify(body),
    })
        .then((response) => console.log(response))
        .catch((error) => console.log(error));
    await page.goto("https://YOUAWESOMEURL");
    await page.click("button");
    await page.click("text=Login");
    await page.fill("#Username", "YOURUSERNAME");
    await page.fill("#Password", "YOURPASSWORD");
    await page.click("#loginButton");
    // await page.click("css=span >> text=Load Test");
    await page.click(
        "#root > div > div > header > ul.nav.navbar-nav.area-tabs > li:nth-child(6) > a >> text=Test"
    );
    await page.waitForSelector("text=SOMETEXTYOUWANTTOCHECKIFTHERE");
    // await page.waitForSelector(`text=SOMEOTHERTEXTYOUWANTTOCHECKIFTHERE`);
    // Another way to check for success
    // await expect(page).toHaveText(`SOMEOTHERTEXTYOUWANTTOCHECKIFTHERE`);
    console.log("test was successful!");
});
于 2020-06-19T03:42:03.280 回答
0

1.19 版本看起来很简单。

test('get respons variable form post in Playwright', async ({ request }) => {
  const responsMy= await request.post(`/repos/${USER}/${REPO}/issues`, {
    data: {
      title: '[Bug] report 1',
      body: 'Bug description',
    }
  });
  expect(responsMy.ok()).toBeTruthy();
}

在https://playwright.dev/docs/test-api-testing上查看更多信息

于 2022-02-19T18:24:57.730 回答