我的routes.rb文件中有以下内容:
resources :businesses do
resources :branches
end
这(正确)生成以下路由:
business_branches GET /businesses/:business_id/branches(.:format) branches#index
POST /businesses/:business_id/branches(.:format) branches#create
new_busines
如何使用计数大小写,何时为sql查询的条件进行搜索。
SELECT Branches.name AS Branches_name, Branches.code AS Branches__code ,
(COUNT(CASE WHEN products_carried_per_branches.carried = 1 THEN 1 END )) AS carried ,
(COUNT(CASE WHEN products_carried_per_branches.carried = 0 THEN 1 END )) AS unCarried
FROM branches INNER JOIN pr
我导入了以下软件包:
import tensorflow as tf
import keras
from keras.models import Sequential, Model
from keras.layers import Conv2D, Flatten, MaxPooling2D, Dense, Input, Reshape, Concatenate, GlobalAveragePooling2D, BatchNormalization, Dropout
from keras.utils import Sequence
import efficientnet.keras as efn
CREATE TABLE banks (
name character varying(49),
id bigint NOT NULL
);
CREATE TABLE branches (
ifsc character varying(11) NOT NULL,
bank_id bigint,
branch character varying(74),
address character varying(195),
city character varying(50),
district character varying(5
这是可行的,它选择多个列:
evaluate SELECTCOLUMNS(branches,
"The branch code", branches[code],
"The branch name", branches[name],
"The branch city", branches[city])
这也适用于:
evaluate branches
order by branches[name]
start at "Co"
但是如果我想将两者结合起
我已经创建了,当我播放它时,我每隔一段时间就会遇到非常大的延迟峰值;当我查看logcat输出时,我可以看到延迟是由垃圾收集器引起的。现在,我已经调整了我的代码,以遵守关于如何制定for循环以不创建垃圾的所有规则,并且我已经在游戏开始之前创建了所有对象。然而,滞后仍然存在。
我看了像Flappy Bird这样的游戏,尽管它很简单,但它没有任何延迟。我做错了什么?
如果需要,我会发布代码。
下面是冲突代码:
branchesSize = Level.branches.size();
for (int x = 0; x < branchesSize; ++x) {
if
我正在尝试编写一个程序,它接受单词并创建一个trie,其中trie的每个节点都是一个包含单个字符的结构。
我有一个函数可以将char*解析成单词(假设char*只包含小写字母)。因为每个单词都是从char*中取出的,所以它被传递给函数addWordOccurrence(const char* word, const int wordLength, struct tNode root)。addWordOccurrence()应该检查单词的第一个字母是否在root.branches[i]中,因为i在循环中递增,检查root.branches的每个可能的索引(对于字母表中的所有小写字母,索引为0-2
我的sql命令如下所示。我正在尝试与两个表建立关系。我需要使用rails助手将查询写到ruby命令。
select *,
(select branch_id
from branches_course_contents
where course_content_id=course_contents.id and branch_id=2) as Sube
from course_contents
where
(select branch_id
from branches_course_contents
where course_content_id
我试图结合我的数据库中的两个查询产生的结果.
q1:
SELECT * FROM werkgevers JOIN werkgevers_branches ON werkgevers.werkgever_id = werkgevers_branches.werkgever_id JOIN plaatsen ON werkgevers.plaats_id = plaatsen.plaats_id JOIN branches ON werkgevers_branches.branche_id = branches.branche_id GROUP BY werkgevers_branches.wer
我想要一个有bno(分支号码)的节点。
struct bank {
struct branch *branches; // to hold the branches of the bank
struct operation_type *optypes; // to hold the transaction types offered by the bank
};
struct branch *getProperBranch(struct bank *banka, int entity) {
while (banka->branches != NULL) {
我有以下疑问:
SELECT users_extra.first_name, users_extra.last_name
FROM (branches, users_extra)
WHERE ((branches.manager_id = users_extra.userid)
OR (branches.sales_manager_id = users_extra.userid)
OR (branches.admin_manager_id = users_extra.userid)
OR (branches.ops_manager_id = users_extra.userid)
OR (br
我有两个模特
分支与Subject
它们通过主语组有多到多的关系
控制台显示以下结果
~/workspace (master) $ rails c
Running via Spring preloader in process 4541
Loading development environment (Rails 4.2.7.1)
irb: warn: can't alias context from irb_context.
2.3.0 :001 > s=Subject.first
Subject Load (0.6ms) SELECT "subjects"
当在cakephp中有或在每个数据中有条件时,我想要在一个案例的和计数的生成数据中实现一个搜索。
SELECT `Branches`.`name` AS `Branches__name` , `Branches`.`code` AS `Branches__code` , (
COUNT(
CASE WHEN `prodCarried`.`carried` =1
THEN 1
END )
) AS `carried` , (
COUNT(
CASE WHEN `prodCarried`.`carried` =0
THEN 1
END )
) AS `unCarried`
FROM `branch
以下是问题所在:
当有人输入domain.com/branches/10时
我需要重定向到domain.com/businesses/1/branches/10
新路
resources :businesses do
resources :branches
end
老路线
match 'branches/:id', to: 'branches#show'
谢谢!
我有一个实体:
public class Branch
{
[Key]
[DatabaseGeneratedAttribute(DatabaseGeneratedOption.Identity)]
public int BranchID { get; set; }
public string Name { get; set; }
public string Address1 { get; set; }
public string Address2 { get; set; }
public string Postcode { get; se