枕梦

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枕梦 2025-02-15 15:36:42

尝试将PLIST文件中的服务名称更改为_ABC-TXTCHAT._TCP。 PLIST文件中的值必须符合Bonjour命名约定,而在构造McNearByserviceadvertiser时,应继续使用abc-txtchat

Try changing the service name in your plist file to _abc-txtchat._tcp. The value in the plist file has to conform to the Bonjour naming convention while you should continue using abc-txtchat while constructing MCNearbyServiceAdvertiser.

McNearByserviceAdvertiser服务器未发布

枕梦 2025-02-15 12:42:34

由于您没有发布任何代码,因此在这种情况下,很难知道如何使用flatmap()。您可以简单地map()在数组上并过滤每个嵌套数组属性。

const array = [{ name: 'Parent Brand 1', childBrands: [{ name: 'Child Brand 1', status: 'active' }, { name: 'Child Brand 2', status: 'discontinued' },] }, { name: 'Parent Brand 2', childBrands: [{ name: 'Child Brand 1', status: 'discontinued' }, { name: 'Child Brand 2', status: 'active' },] }];

const filtered = array.map(parent => ({
  ...parent,
  childBrands: parent.childBrands.filter(child => child.status === 'active')
}));

console.log(filtered)

Since you haven't posted any code it's hard to know how you would be using flatMap() in this case. You can simply map() over the array and filter each nested array property.

const array = [{ name: 'Parent Brand 1', childBrands: [{ name: 'Child Brand 1', status: 'active' }, { name: 'Child Brand 2', status: 'discontinued' },] }, { name: 'Parent Brand 2', childBrands: [{ name: 'Child Brand 1', status: 'discontinued' }, { name: 'Child Brand 2', status: 'active' },] }];

const filtered = array.map(parent => ({
  ...parent,
  childBrands: parent.childBrands.filter(child => child.status === 'active')
}));

console.log(filtered)

如何在子元素上执行操作并退还父母?

枕梦 2025-02-15 10:43:32
void LoadFile(vector<vector<string>> courseInfo) { ... }

这会创建向量的副本,因为您按值将其传递。然后,您将所有数据存储在副本中,并在功能末尾将副本销毁。您绝对不会修改原始矢量。将其更改为

void LoadFile(vector<vector<string>> &courseInfo) { ... }
void LoadFile(vector<vector<string>> courseInfo) { ... }

This creates a copy of the vector because you pass it by value. You then store all the data in the copy and at the end of the function the copy is destroyed. At no point do you modify the original vector. Change it to

void LoadFile(vector<vector<string>> &courseInfo) { ... }

在C&#x2B;&#x2B中使用2D向量的数据使用来自2D矢量的数据。

枕梦 2025-02-15 00:43:01

不客气:)

默认情况下,Docker将内部网络中的服务连接。要重新使用其他服务,您需要使用服务名称(或Aliases ...),Localhost将无法使用。

第二期发布的内容与RabbitMQ连接无关,您的MYTB_1容器无法访问您的数据卷映射。该服务需要有足够的权利来写入映射的目录

version: '2.2'
services:
  mytb:
    privileged: true
    restart: always
    depends_on: 
      - rabbitmq
    image: "thingsboard/tb-postgres"
    ports:
      - "9091:9090"
      - "1883:1883"
      - "7070:7070"
      - "5683-5688:5683-5688/udp"
    environment:
      TB_QUEUE_TYPE: rabbitmq
      TB_QUEUE_RABBIT_MQ_USERNAME: test
      TB_QUEUE_RABBIT_MQ_PASSWORD: test123369
      # Use the rabbitmq service name as a host instead of localhost
      TB_QUEUE_RABBIT_MQ_HOST: rabbitmq 
      TB_QUEUE_RABBIT_MQ_PORT: 5672
    volumes:        
      - ./.mytb-data:/data
      - ./.mytb-logs:/var/log/thingsboard
      # make sure that the service has sufficient rights to
      # write to the mapped directories
  rabbitmq:
    ... (rest of file left out)

You're welcome :)

By default, docker connects services within an internal network. To refence other services, you need to use the service name (or aliases...), localhost will not work.

Second issue posted doesn't have something to to with the rabbitmq connection, your mytb_1 container cannot access your data volume mapping. The service needs to have sufficient rights to write to the mapped directories

version: '2.2'
services:
  mytb:
    privileged: true
    restart: always
    depends_on: 
      - rabbitmq
    image: "thingsboard/tb-postgres"
    ports:
      - "9091:9090"
      - "1883:1883"
      - "7070:7070"
      - "5683-5688:5683-5688/udp"
    environment:
      TB_QUEUE_TYPE: rabbitmq
      TB_QUEUE_RABBIT_MQ_USERNAME: test
      TB_QUEUE_RABBIT_MQ_PASSWORD: test123369
      # Use the rabbitmq service name as a host instead of localhost
      TB_QUEUE_RABBIT_MQ_HOST: rabbitmq 
      TB_QUEUE_RABBIT_MQ_PORT: 5672
    volumes:        
      - ./.mytb-data:/data
      - ./.mytb-logs:/var/log/thingsboard
      # make sure that the service has sufficient rights to
      # write to the mapped directories
  rabbitmq:
    ... (rest of file left out)

如何模拟物构成东西板&#x2B;兔子?

枕梦 2025-02-14 20:56:40

如果 - 属性 statement utate 这样(出于多种原因);相反,您可以使用case_when

data <-
  data %>% 
  mutate(second.indicator = case_when((il.count.description + un.count.description) > (elle.count.description + une.count.description) ~ -1,
                                      (il.count.description + un.count.description) == (elle.count.description + une.count.description) ~ 0,
                                      T ~ 1
                                      )

You cannot use if-statements inside mutate like that (for multiple reasons); instead you could use case_when:

data <-
  data %>% 
  mutate(second.indicator = case_when((il.count.description + un.count.description) > (elle.count.description + une.count.description) ~ -1,
                                      (il.count.description + un.count.description) == (elle.count.description + une.count.description) ~ 0,
                                      T ~ 1
                                      )

否则如果语句创建新列

枕梦 2025-02-14 00:53:41

似乎DOT NET核心,实体框架(EF)和C#在构建复杂查询(旧数据)的情况下有多个LINQ的问题。最终,我通过简单连接的多个列表创建了数据。
所有先前的陈述都出现为&lt;&gt; h__ transparentIdentifier,在与EF打交道时似乎行为的行为。但是,鉴于本地列表,系统丢弃了上述错误,这是可以理解的,因为SQL编译器对列表不了解,我希望LINQ能够作为一组次级征服而不是尝试形成单个查询。

It seems that dot net core, Entity Framework (EF) and C# has an issue with multiple LINQ from calls in building up a complex query(legacy data). I eventually created the data via multiple lists with simple joins.
All previous from statements appear as <>h__TransparentIdentifier and when dealing with EF seemed to behave. but given a local list the system threw the above error, which is understandable as the SQL compiler has no Knowledge of the List, I hoped that the LINQ would compile as a group of subqueries rather than trying to form a single query.

linq和核心简单从句和投影问题

枕梦 2025-02-14 00:14:39

我当前的应用具有appconfig类,该类别扩展 norfollow noreferrer“> sentaritedwidgetwidget 可以使实例可用的实例在使用中。

您可以尝试使用

class AppConfig extends InheritedWidget {

    static AppConfig? of(BuildContext context) {
        initializeFirebase();
        return context.dependOnInheritedWidgetOfExactType<AppConfig>();
      }
    
    static Future<void> initializeFirebase() async {
        try {
          await Firebase.initializeApp();
        } catch (e) {}
      }
}

Main()中的AppConfig()初始化

My current app has AppConfig class that extends InheritedWidget that keeps instance available when in use.

You can try with

class AppConfig extends InheritedWidget {

    static AppConfig? of(BuildContext context) {
        initializeFirebase();
        return context.dependOnInheritedWidgetOfExactType<AppConfig>();
      }
    
    static Future<void> initializeFirebase() async {
        try {
          await Firebase.initializeApp();
        } catch (e) {}
      }
}

You need to initialise the AppConfig() in main()

如何将FlutterFire用于模拟器演示项目?

枕梦 2025-02-13 17:07:31

这是由于在字符串开始时插入了Bom。我必须指定正确的编码字符集来解决此问题。
请检查 u'\ ufeff'

print(repr(finalSql))

有关详细信息,

insert into OTBI_MANUFACTURERS (\ufeffMANUFACTURER,REGISTRY_ID,MFR_DESCRIPTION,STATUS,PARTY_TYPE,CITY,STATE,POSTAL_CODE,PRIMARY_PHONE_NUMBER,PRIMARY_PHONE_EXTENSION,WEBSITE,CONTACT_NAME,GENERAL_INQUIRY_EMAIL,MANUFACTURER_NAME_LONG,MANUFACTURER_ADDRESS,COUNTRY,ADDRESS1,SUPPLIER_NAME,SUPPLIER_EMAIL,SUPPLIER_PHONE_NUMBER,ADDRESS2) values (:1,:2,:3,:4,:5,:6,:7,:8,:9,:10,:11,:12,:13,:14,:15,:16,:17,:18,:19,:20,:21)'

It was due to BOM being inserted at the start of the string. I had to specify the correct encoding character set to solve this.
For details check u'\ufeff' in Python string

print(repr(finalSql))

Output:

insert into OTBI_MANUFACTURERS (\ufeffMANUFACTURER,REGISTRY_ID,MFR_DESCRIPTION,STATUS,PARTY_TYPE,CITY,STATE,POSTAL_CODE,PRIMARY_PHONE_NUMBER,PRIMARY_PHONE_EXTENSION,WEBSITE,CONTACT_NAME,GENERAL_INQUIRY_EMAIL,MANUFACTURER_NAME_LONG,MANUFACTURER_ADDRESS,COUNTRY,ADDRESS1,SUPPLIER_NAME,SUPPLIER_EMAIL,SUPPLIER_PHONE_NUMBER,ADDRESS2) values (:1,:2,:3,:4,:5,:6,:7,:8,:9,:10,:11,:12,:13,:14,:15,:16,:17,:18,:19,:20,:21)'

无法求解cx_oracle.databaseerror:ora-00911:无效字符

枕梦 2025-02-13 12:46:22

有三个部分:1)找到零,2)找到平均值,3)填充发现的零。因此:

/****
 * in-place fill zeros with the mean of the surrounding neighborhoods
 ***/
void fillHoles(Mat gray){    
    // find the zeros
    Mat mask = (gray == 0);
    
    // find the mean with filter2d
    Mat kernel = (Mat_<double>(3,3) << 
    1/8, 1/8, 1/8
    1/8, 0  , 1/8
    1/8, 1/8, 1/8
    );
    Mat avg;
    cv::filter2d(gray, avg, CV_8U, kernel)
    
    // then fill the zeros, only where indicated by `mask`
    cv::bitwise_or(gray, avg, gray, mask);

}

注释我只是意识到这显然是在获得平均值,而不是非零的平均值。对于该操作,您可能需要执行两个过滤器,一个用于总和,一个用于非零计数,然后将两个过滤器划分为:

// find the neighborhood sum with filter2d
Mat kernel = (Mat_<double>(3,3) << 
1, 1, 1
1, 0, 1
1, 1, 1
);
Mat sums;
cv::filter2d(gray, sums, CV_64F, kernel);

// find the neighborhood count with filter2d
Mat counts;
cv::filter2d(gray!=0, counts, CV_64F, kernel);    
counts /= 255; // because gray!=0 returns 255 where true

// force counts to 1 if 0, so we can divide later
cv::max(counts, 1, counts);

Mat out;
cv::divide(sums, counts, out);

out.convertTo(gray, CV_8U);

There are three parts: 1) find the zeros, 2) find the mean, and 3) fill the found zeros with mean. So:

/****
 * in-place fill zeros with the mean of the surrounding neighborhoods
 ***/
void fillHoles(Mat gray){    
    // find the zeros
    Mat mask = (gray == 0);
    
    // find the mean with filter2d
    Mat kernel = (Mat_<double>(3,3) << 
    1/8, 1/8, 1/8
    1/8, 0  , 1/8
    1/8, 1/8, 1/8
    );
    Mat avg;
    cv::filter2d(gray, avg, CV_8U, kernel)
    
    // then fill the zeros, only where indicated by `mask`
    cv::bitwise_or(gray, avg, gray, mask);

}

Note I just realize that this is plainly taking the average, not the non-zero average. For that operation, you might want to do two filters, one for the sum, one for the non-zero counts, then divide the two:

// find the neighborhood sum with filter2d
Mat kernel = (Mat_<double>(3,3) << 
1, 1, 1
1, 0, 1
1, 1, 1
);
Mat sums;
cv::filter2d(gray, sums, CV_64F, kernel);

// find the neighborhood count with filter2d
Mat counts;
cv::filter2d(gray!=0, counts, CV_64F, kernel);    
counts /= 255; // because gray!=0 returns 255 where true

// force counts to 1 if 0, so we can divide later
cv::max(counts, 1, counts);

Mat out;
cv::divide(sums, counts, out);

out.convertTo(gray, CV_8U);

OPENCV C&#x2B;&#x2B;

枕梦 2025-02-12 01:25:55

可以使用data(REGEX 在这里) ,知道线程名称在括号内,所以

%{DATE_EU:date} %{TIME:time} %{ISO8601_TIMEZONE:zone} \[%{LOGLEVEL:loglevel} \] \[%{DATA:thread_name}\].*

It is possible to use DATA (regex .*?) instead of GREEDYDATA (regex .*) (code here), knowing that the thread name is inside brackets, so something like

%{DATE_EU:date} %{TIME:time} %{ISO8601_TIMEZONE:zone} \[%{LOGLEVEL:loglevel} \] \[%{DATA:thread_name}\].*

线程名称的Grok模式(Logstash 8.2.0)

枕梦 2025-02-11 22:16:31

搜索列表

您可以使用以下代码:

idx = np.searchsorted(df.index, num)

df.index[max(0, idx)]

示例

数据集为:

df = pd.DataFrame({'Time': range(10), 'Alt': range(10)}, index=range(0,20, 2))

Output:

    Time  Alt
0      0    0
2      1    1
4      2    2
6      3    3
8      4    4
10     5    5
12     6    6
14     7    7
16     8    8
18     9    9

如果

num = 7
idx = np.searchsorted(df.index, num)
df.index[max(0, idx)]

Output:

8

要在输入的索引号下方找到索引,则更改:

df.index[max(0, idx)]

8

df.index[max(0, idx-1)]

从上面的示例输出为6,而不是

和为了良好的衡量,这是一个可以处理这一切的函数:

def next_index(df, num):
    # Returns the next index from num in the df
    idx = np.searchsorted(df.index, num)
    return df.index[max(0, idx)]

Searchsorted

You can use the following code:

idx = np.searchsorted(df.index, num)

df.index[max(0, idx)]

Example

Dataset is:

df = pd.DataFrame({'Time': range(10), 'Alt': range(10)}, index=range(0,20, 2))

Output:

    Time  Alt
0      0    0
2      1    1
4      2    2
6      3    3
8      4    4
10     5    5
12     6    6
14     7    7
16     8    8
18     9    9

Then

num = 7
idx = np.searchsorted(df.index, num)
df.index[max(0, idx)]

Output:

8

If you want to find the index below the index number you enter, then change:

df.index[max(0, idx)]

to

df.index[max(0, idx-1)]

Output from the above example would be 6 instead of 8

And just for good measure, here is a function that can handle this all:

def next_index(df, num):
    # Returns the next index from num in the df
    idx = np.searchsorted(df.index, num)
    return df.index[max(0, idx)]

如何以最接近熊猫的一定值获取数据

枕梦 2025-02-11 02:36:20

您可以使用 string.join()

Console.WriteLine(string.Join("<span>l</span>", words));

而不是

//No need to build resultString as well. That mean you can eliminate for loop
Console.WriteLine(resultString);


注意:这只是与您的问题保持一致的另一种方式。 @fuzzybear或@ash提出的解决方案比我建议的更优雅。

You can do it using string.Join()

Console.WriteLine(string.Join("<span>l</span>", words));

Instead of

//No need to build resultString as well. That mean you can eliminate for loop
Console.WriteLine(resultString);

.Try online


Note: This is just an alternate way aligned to your question. The solution proposed by @fuzzybear or @ASh are more elegant than what I suggested.

C#在字符串中查找字符串的出现,并在匹配的字符上包裹文本

枕梦 2025-02-10 18:46:38

最后,我不得不添加两个额外的滴答:

ROM DISK = '''

* @backupPath + @backupFile + ''' WITH STANDBY = N''D:\TelcorLogDump\ROLLBACK_UNDO_Telcor.BAK

'''

I had to add two extra ticks at the end:

ROM DISK = '''

* @backupPath + @backupFile + ''' WITH STANDBY = N''D:\TelcorLogDump\ROLLBACK_UNDO_Telcor.BAK

'''

从磁盘还原多个事务日志备份

枕梦 2025-02-10 03:37:44
order.cartItems.forEach(async (item) => {
      const result = await db
        .collection("items")
        .updateOne(
          { _id: parseInt(item.itemId) },
          { $inc: { numInStock: -parseInt(item.quantity) } }
        );
    });

然后恢复numinstock

result.cartItems.forEach(async (item) => {
      const result = await db
        .collection("items")
        .updateOne(
          { _id: parseInt(item.itemId) },
          { $inc: { numInStock: parseInt(item.quantity) } }
        );
    });
order.cartItems.forEach(async (item) => {
      const result = await db
        .collection("items")
        .updateOne(
          { _id: parseInt(item.itemId) },
          { $inc: { numInStock: -parseInt(item.quantity) } }
        );
    });

then to restore the numInStock

result.cartItems.forEach(async (item) => {
      const result = await db
        .collection("items")
        .updateOne(
          { _id: parseInt(item.itemId) },
          { $inc: { numInStock: parseInt(item.quantity) } }
        );
    });

如何通过嵌套文档作为JSON主体迭代,并更新集合中的每个文档

枕梦 2025-02-10 02:30:39

由于usestate是异步直接替换状态不是一个好主意。因此,最好访问先前的状态并设置新的更新值。

for ex:setState(((prev)=&gt; [... prev,... result]);
//结果是对象数组

Since useState is async direct replacement of state is not a good idea. so it is better to access the previous state and set new updated value.

for ex: setState((prev)=>[...prev, ...result]);
//result is array of objects

USESTATE SET方法不是立即反映更改

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