我正在开发新闻应用程序，但从Gradle控制台中收到以下错误
(androidx.core:core:1.0.1) and classes.jar (com.android.support:support-compat:28.0.0)
Duplicate class android.support.v4.app.INotificationSideChannel$Stub found in modules classes.jar (androidx.core:core:1.0.1) and classes.jar (com.android.support:support-compat:28.0.0)
Duplicate class android.support.v4.app.INotificationSideChannel$Stub$Proxy found in modules classes.jar (androidx.core:core:1.0.1) and classes.jar (com.android.support:support-compat:28.0.0)
Duplicate class android.support.v4.graphics.drawable.IconCompatParcelizer found in modules classes.jar (androidx.core:core:1.0.1) and classes.jar (com.android.support:support-compat:28.0.0)
Duplicate class android.support.v4.os.IResultReceiver found in modules classes.jar (androidx.core:core:1.0.1) and classes.jar (com.android.support:support-compat:28.0.0)
Duplicate class android.support.v4.os.IResultReceiver$Stub found in modules classes.jar (androidx.core:core:1.0.1) and classes.jar (com.android.support:support-compat:28.0.0)
Duplicate class android.support.v4.os.IResultReceiver$Stub$Proxy found in modules classes.jar (androidx.core:core:1.0.1) and classes.jar (com.android.support:support-compat:28.0.0)
Duplicate class android.support.v4.os.ResultReceiver found in modules classes.jar (androidx.core:core:1.0.1) and …在 Termux (my_distro) 中：
$ pkg show openssl
Package: openssl
Version: 3.0.1-1
Maintainer: @termux
Installed-Size: 6648 kB
Depends: ca-certificates, zlib
Conflicts: libcurl (<< 7.61.0-1)
Breaks: openssl-tool (<< 1.1.1b-1), openssl-dev
Replaces: openssl-tool (<< 1.1.1b-1), openssl-dev
openssl 安装的软件包：
openssh/stable 8.8p1-3 aarch64 [upgradable from: 8.8p1-1]
openssl-tool/stable 3.0.1-1 aarch64 [upgradable from: 1.1.1m-2]
openssl/stable 3.0.1-1 aarch64 [upgradable from: 1.1.1m-2]
Node.js的安装：
$ pkg install nodejs-lts
Setting up nodejs-lts (16.14.0-2) ...
CANNOT LINK EXECUTABLE "node": library "libcrypto.so.3" not found
Aborted
dpkg: error processing package nodejs-lts (--configure):
installed …我有这个代码：
from pydantic import BaseModel, constr
DeptNumber = constr(min_length=6, max_length=6)
class MyStuff(BaseModel):
dept: DeptNumber
ms = MyStuff(dept = "123456")
deptnr.py:6：错误：变量“deptnr.DeptNumber”作为类型无效
deptnr.py:6：注意：请参阅https://mypy.readthedocs.io/en/latest/common_issues.html#variables-vs -类型别名
提供的链接似乎并没有真正解决我的问题（我没有使用Type）。
不管有没有这个都会发生这种情况mypy.ini：
[mypy]
plugins = pydantic.mypy
[pydantic-mypy]
init_typed = true
最初我在 Pydantic 中也遇到了choice如下错误，但我通过使用 Python 解决了这个问题Literal。
DIR = choice(["North", "East", "South", "West"])
我需要改变什么才能让 mypy 对 Pydantic 感到满意constr？
“werkzeug.routing.BuildError：无法为端点'success'构建url。您是否忘记指定值['name']？”
如何解决这个错误？我尝试了很多事情，但无法自己解决。
http_methods.py：
from flask import Flask, redirect, url_for, request
app = Flask(__name__)
@app.route('/success/<name>')
def success(name):
return 'welcome %s' % name
@app.route('/login', methods=['POST', 'GET'])
def login():
if request.method == 'POST':
user = request.form['nm']
return redirect(url_for('success', name = user))
else:
user = request.args.get('nm')
return redirect(url_for('success', name = user))
if __name__ == "__main__":
app.run(debug=True)
我在登录方法中遇到错误，未指定 ['name'] 的值；如何解决这个错误？
登录.html：
<!DOCTYPE html>
<html lang="en">
<body>
<form action="http://127.0.0.1:5000/login" method="POST">
<p>Enter name:</p>
<p><input type="text" name="nm" value="nm"/></p>
<p><input type="submit" value="submit"/></p>
</form>
</body>
</html>
输出：
File "/Users/chirag.kanzariya/pythonprojects/v_python/lib/python3.7/site-packages/flask/helpers.py", line …以下是我的api.yml：
openapi: 3.0.0
info:
title: 'AutoDelievery API'
description: 'AutoDelievery API Documentation'
termsOfService: 'https://www.yopmail.com'
contact:
name: yopmail
url: 'http://www.yopmail.com'
email: contactus@yopmail.com
license:
name: yopmail
url: 'http://www.yopmail.com/licenses/yopmail-AutoDelievery-Licence.html'
version: 1.0.0-oas3
servers:
-
url: 'http://localhost:5055/'
tags:
-
name: AutoDelievery
description: 'AutoDelievery product related APIs'
externalDocs:
description: 'Documentation:'
url: 'http://localhost:5055/external/doc'
paths:
/autoDelievery:
post:
tags: [autoDelievery]
summary: 'Creates a new autoDelievery record'
operationId: insertSubscription
requestBody: {description: 'autoDelievery request body', content: {application/json: {schema: {$ref: '#/components/schemas/AutoDelievery'}}}}
responses: {'200': {description: 'successful operation', content: {application/json: {schema: {$ref: '#/components/schemas/AutoDelievery'}}}}, '400': {description: …我正在本地 PC 上设置一个新的 kubernetes 设置，其规格如下。在尝试启动 Kubernetes 集群时，我遇到了一些问题。需要您的投入。
操作系统版本：Linux server.cent.com 3.10.0-123.el7.x86_64 #1 SMP Mon Jun 3在此处输入代码`0 12:09:22 UTC 2014 x86_64 x86_64 x86_64 GNU/Linux
Docker版本：Docker版本1.13.1，内部版本07f3374/1.13.1
[root@server ~]# rpm -qa |grep -i kube
kubectl-1.13.2-0.x86_64
kubernetes-cni-0.6.0-0.x86_64
kubeadm-1.13.2-0.x86_64
kubelet-1.13.2-0.x86_64
面临的问题是：
[root@server ~]# kubeadm init --apiserver-advertise-address=192.168.203.154 --pod-network-cidr=10.244.0.0/16
[kubelet-check] Initial timeout of 40s passed.
Unfortunately, an error has occurred:
timed out waiting for the condition
This error is likely caused by:
- The kubelet is not running
- The kubelet is unhealthy due to a misconfiguration of …我觉得必须有更好的方法来计算事件,而不是在Linux中的perl,shell中编写sub.
#/usr/bin/perl -w
use strict;
return 1 unless $0 eq __FILE__;
main() if $0 eq __FILE__;
sub main{
my $str = "ru8xysyyyyyyysss6s5s";
my $char = "y";
my $count = count_occurrence($str, $char);
print "count<$count> of <$char> in <$str>\n";
}
sub count_occurrence{
my ($str, $char) = @_;
my $len = length($str);
$str =~ s/$char//g;
my $len_new = length($str);
my $count = $len - $len_new;
return $count;
}
我在raku rest / API应用程序上动态加载模块。
我想避免每次对模块进行任何修改时服务器都重新启动。
有没有一种方法可以在您每次调用require时重新加载raku模块？
try require ::('Foo');
在Java中，我们如何在不使用签名的情况下验证给定的字符串是否是JWT令牌？
我在用
try {
return (new JwtConsumerBuilder()).setVerificationKey(SECRET_KEY).build().processToClaims(token);
} catch (InvalidJwtException var4) {
throw new IOException("Failed to parse");
}
这工作正常，但我想在没有SECRET_KEY.
我只是想验证它是否是 JWT 令牌。
我想使用 Pydantic 来验证对象中的字段，但似乎验证仅在我创建实例时发生，而不是在我稍后修改字段时发生。
from pydantic import BaseModel, validator
class MyStuff(BaseModel):
name: str
@validator("name")
def ascii(cls, v):
assert v.isalpha() and v.isascii(), "must be ASCII letters only"
return v
# ms = MyStuff(name = "me@example.com") # fails as expected
ms = MyStuff(name = "me")
ms.name = "me@example.com"
print(ms.name) # prints me@example.com
MyStuff在上面的示例中，当我按照预期在创建时尝试传递无效值时，Pydantic 会抱怨。
但是当我之后修改该字段时，Pydantic 并没有抱怨。这是预期的吗？或者我如何让 Pydantic 在分配字段时也运行验证器？
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