Using Playwright and Python, how can I expect for count bigger or smaller than?

使用Playwright和Python，我怎么能期望计数大于或小于？

For example, this code expect for count of 2.

例如，此代码期望计数为2。

How do I achieve count >= 2 (Only bigger)

如何实现计数>=2（仅更大）

expect(self.page.locator('MyLocator')).to_have_count(2, timeout=20 * 1000)

Playwright doesn't have options to use bigger/smaller in toHaveCount/to_have_count.

Playwright没有在toHaveCount/to_have_count中使用更大/更小的选项。

I guess you can rewrite it to Python easily.

我想你可以很容易地把它重写成Python。

const count = await page.locator("MyLocator").count();
expect(count).toBeGreaterThan(2);
expect(count).toBeLessThanOrEqual(2);

The only limitation of this approach is that you won't have expect retry feature because you check the int value, not a Locator. So you might wanna check for the visibility of the Locator first.

这种方法的唯一限制是，您不会期望有重试功能，因为您检查的是int值，而不是Locator。因此，您可能需要先检查定位器的可见性。

const locator = page.locator("MyLocator");
await expect(locator.last()).toBeVisible(); // or first()
const count = await locator.count();
expect(count).toBeGreaterThan(2);
expect(count).toBeLessThanOrEqual(2);

Also, PW for NodeJS has the ability to create custom matches, but I'm not sure it's an option for Python.

此外，PW for NodeJS能够创建自定义匹配，但我不确定这是否是Python的选项。

This doesn't seem to be possible in the current Python Playwright API, but you could use wait_for_function as a workaround:

在当前的Python Playwright API中，这似乎是不可能的，但您可以使用wait_for_formation作为解决方法：

page.wait_for_function("document.querySelectorAll('.foo').length > 2")

This is web-first and will wait for the predicate, but the error message once it throws won't be as clear as the expect failure.

这是web优先的，将等待谓词，但一旦抛出错误消息，就不会像预期的失败那样清晰。

If the count is immediately available and you don't need to wait for the predicate to be true, assert is useful to mention as another possibility:

如果计数立即可用，并且您不需要等待谓词为true，那么断言作为另一种可能性是有用的：

assert page.locator(".foo").count() > 2

If you're using unittest, you can replace assert with, for example

如果您使用的是unittest，您可以用替换assert，例如

self.assertGreaterEqual(page.locator(".foo").count(), 2)
self.assertGreater(page.locator(".foo").count(), 2)  # or

Yet another workaround if you're dealing only with small numbers of elements:

如果您只处理少量元素，还有另一种解决方法：

page.locator(".foo")
expect(loc).not_to_have_count(0)
expect(loc).not_to_have_count(1)
expect(loc).not_to_have_count(2)

Here, we ensure the count is greater than 2 by process of elimination. It's impossible for count to be less than 0, so we need not include that.

在这里，我们通过消除过程来确保计数大于2。计数不可能小于0，所以我们不需要包括它。

You could do this for less than as well, but not as easily, and using a reasonable assumption that there'll never be more than, say, 20 or 50 elements in any conceivable run:

你可以用不到的时间，但不那么容易，并使用一个合理的假设，即在任何可能的运行中，永远不会有超过20或50个元素：

loc = page.locator(".foo")
for i in range(2, 20):
    expect(loc).not_to_have_count(i)

This ensures the count is 0 or 1, or greater than some reasonably high upper-bound.

这样可以确保计数为0或1，或者大于某个合理的上限。

Use not_to_have_count

使用not_to_have_count

expect(self.page.locator('MyLocator')).not_to_have_count(2, timeout=20 * 1000)

Just found this workaround and it worked like a charm:

刚刚找到了这个变通方法，它就像一个魅力：

"You can do expect(await locator.count()).toBeGreatedThen() as a workaround for now. It won't be waiting until the number is greater than, it will just assert it immediately, but it looks like it might work for you."

“你可以期待（wait-locator.count（））.toBeGreatedThen（）作为目前的解决方法。它不会等到数字大于，它只会立即断言，但看起来它可能对你有用。”

in https://github.com/microsoft/playwright/issues/8725

在里面https://github.com/microsoft/playwright/issues/8725

from playwright.sync_api import sync_playwright

with sync_playwright() as p:
    browser = p.chromium.launch()
    page = browser.new_page()
    page.goto('https://example.com')  # Replace with your URL
    locator = page.locator('MyLocator')

    # Wait for the element to appear within a timeout
    locator.wait_for_selector(timeout=20000)

    # Get the count of elements matching the locator
    count = locator.count()

    # Check if count is greater than 2 or less than or equal to 2
    if count > 2:
        print(f"Count is greater than 2: {count}")
    elif count <= 2:
        print(f"Count is less than or equal to 2: {count}")

    browser.close()

use this and let me have your respect if it works good.
'''ERFAN MOGHIS'''

用这个，如果效果好的话，请尊重我ERFAN MOGHIS