- html - 出于某种原因,IE8 对我的 Sass 文件中继承的 html5 CSS 不友好?
- JMeter 在响应断言中使用 span 标签的问题
- html - 在 :hover and :active? 上具有不同效果的 CSS 动画
- html - 相对于居中的 html 内容固定的 CSS 重复背景?
处理多对多关系时,如何为选择列表实现循环级联逻辑?
例如,我创建了一个简单的测试应用程序,该程序可以跟踪书籍和作者。
(这比我的实际业务场景更简单,并且更清楚地显示了问题。)
主页包含:
CREATE table "BOOK" (
"ID" INTEGER GENERATED ALWAYS AS IDENTITY NOT NULL ENABLE,
"TITLE" VARCHAR2(100) NOT NULL ENABLE,
constraint "BOOK_CK" check ("TITLE"<>''),
constraint "BOOK_PK" primary key ("ID"),
constraint "BOOK_UK1" unique ("TITLE")
)
/
CREATE table "AUTHOR" (
"ID" INTEGER GENERATED ALWAYS AS IDENTITY NOT NULL ENABLE,
"NAME" VARCHAR2(100) NOT NULL ENABLE,
constraint "AUTHOR_CK" check ("NAME"<>''),
constraint "AUTHOR_PK" primary key ("ID"),
constraint "AUTHOR_UK1" unique ("NAME")
)
/
INSERT INTO BOOK (TITLE) VALUES ('BOOK01');
INSERT INTO BOOK (TITLE) VALUES ('BOOK02');
INSERT INTO BOOK (TITLE) VALUES ('BOOK03');
INSERT INTO BOOK (TITLE) VALUES ('BOOK04');
INSERT INTO BOOK (TITLE) VALUES ('BOOK05');
INSERT INTO BOOK (TITLE) VALUES ('BOOK06');
INSERT INTO BOOK (TITLE) VALUES ('BOOK07');
INSERT INTO BOOK (TITLE) VALUES ('BOOK08');
INSERT INTO BOOK (TITLE) VALUES ('BOOK09');
INSERT INTO BOOK (TITLE) VALUES ('BOOK10');
INSERT INTO BOOK (TITLE) VALUES ('BOOK11');
INSERT INTO BOOK (TITLE) VALUES ('BOOK12');
INSERT INTO BOOK (TITLE) VALUES ('BOOK13');
INSERT INTO BOOK (TITLE) VALUES ('BOOK14');
INSERT INTO BOOK (TITLE) VALUES ('BOOK15');
INSERT INTO BOOK (TITLE) VALUES ('BOOK16');
INSERT INTO BOOK (TITLE) VALUES ('BOOK17');
INSERT INTO BOOK (TITLE) VALUES ('BOOK18');
INSERT INTO BOOK (TITLE) VALUES ('BOOK19');
INSERT INTO BOOK (TITLE) VALUES ('BOOK20');
INSERT INTO AUTHOR (NAME) VALUES ('AUTHOR01');
INSERT INTO AUTHOR (NAME) VALUES ('AUTHOR02');
INSERT INTO AUTHOR (NAME) VALUES ('AUTHOR03');
INSERT INTO AUTHOR (NAME) VALUES ('AUTHOR04');
INSERT INTO AUTHOR (NAME) VALUES ('AUTHOR05');
INSERT INTO AUTHOR (NAME) VALUES ('AUTHOR06');
INSERT INTO AUTHOR (NAME) VALUES ('AUTHOR07');
INSERT INTO AUTHOR (NAME) VALUES ('AUTHOR08');
INSERT INTO AUTHOR (NAME) VALUES ('AUTHOR09');
INSERT INTO AUTHOR (NAME) VALUES ('AUTHOR10');
INSERT INTO AUTHOR (NAME) VALUES ('AUTHOR11');
INSERT INTO AUTHOR (NAME) VALUES ('AUTHOR12');
INSERT INTO AUTHOR (NAME) VALUES ('AUTHOR13');
INSERT INTO AUTHOR (NAME) VALUES ('AUTHOR14');
INSERT INTO AUTHOR (NAME) VALUES ('AUTHOR15');
INSERT INTO AUTHOR (NAME) VALUES ('AUTHOR16');
INSERT INTO AUTHOR (NAME) VALUES ('AUTHOR17');
INSERT INTO AUTHOR (NAME) VALUES ('AUTHOR18');
INSERT INTO AUTHOR (NAME) VALUES ('AUTHOR19');
INSERT INTO AUTHOR (NAME) VALUES ('AUTHOR20');
CREATE table "BOOK_AUTHOR" (
"ID" INTEGER GENERATED ALWAYS AS IDENTITY NOT NULL ENABLE,
"BOOK_ID" INTEGER NOT NULL ENABLE,
"AUTHOR_ID" INTEGER NOT NULL ENABLE,
constraint "BOOK_AUTHOR_PK" primary key ("ID"),
constraint "BOOK_AUTHOR_UK1" unique ("BOOK_ID","AUTHOR_ID")
)
/
ALTER TABLE BOOK_AUTHOR ADD FOREIGN KEY (BOOK_ID)
REFERENCES BOOK (ID) ENABLE
/
ALTER TABLE BOOK_AUTHOR ADD FOREIGN KEY (AUTHOR_ID)
REFERENCES AUTHOR (ID) ENABLE
/
CREATE OR REPLACE VIEW "VW_BOOK_AUTHOR" AS
SELECT ba.ID,
ba.BOOK_ID,
b.TITLE BOOK_TITLE,
ba.AUTHOR_ID,
a.NAME AUTHOR_NAME
FROM BOOK_AUTHOR ba
LEFT JOIN BOOK b on b.ID = ba.BOOK_ID
LEFT JOIN AUTHOR a on a.ID = ba.AUTHOR_ID
/
INSERT INTO BOOK_AUTHOR (BOOK_ID,AUTHOR_ID) SELECT b.ID, a.ID FROM BOOK b, AUTHOR a WHERE b.TITLE='BOOK01' AND a.NAME='AUTHOR01';
INSERT INTO BOOK_AUTHOR (BOOK_ID,AUTHOR_ID) SELECT b.ID, a.ID FROM BOOK b, AUTHOR a WHERE b.TITLE='BOOK02' AND a.NAME='AUTHOR02';
INSERT INTO BOOK_AUTHOR (BOOK_ID,AUTHOR_ID) SELECT b.ID, a.ID FROM BOOK b, AUTHOR a WHERE b.TITLE='BOOK03' AND a.NAME='AUTHOR03';
INSERT INTO BOOK_AUTHOR (BOOK_ID,AUTHOR_ID) SELECT b.ID, a.ID FROM BOOK b, AUTHOR a WHERE b.TITLE='BOOK04' AND a.NAME='AUTHOR04';
INSERT INTO BOOK_AUTHOR (BOOK_ID,AUTHOR_ID) SELECT b.ID, a.ID FROM BOOK b, AUTHOR a WHERE b.TITLE='BOOK05' AND a.NAME='AUTHOR05';
INSERT INTO BOOK_AUTHOR (BOOK_ID,AUTHOR_ID) SELECT b.ID, a.ID FROM BOOK b, AUTHOR a WHERE b.TITLE='BOOK06' AND a.NAME='AUTHOR06';
INSERT INTO BOOK_AUTHOR (BOOK_ID,AUTHOR_ID) SELECT b.ID, a.ID FROM BOOK b, AUTHOR a WHERE b.TITLE='BOOK07' AND a.NAME='AUTHOR07';
INSERT INTO BOOK_AUTHOR (BOOK_ID,AUTHOR_ID) SELECT b.ID, a.ID FROM BOOK b, AUTHOR a WHERE b.TITLE='BOOK08' AND a.NAME='AUTHOR08';
INSERT INTO BOOK_AUTHOR (BOOK_ID,AUTHOR_ID) SELECT b.ID, a.ID FROM BOOK b, AUTHOR a WHERE b.TITLE='BOOK09' AND a.NAME='AUTHOR09';
INSERT INTO BOOK_AUTHOR (BOOK_ID,AUTHOR_ID) SELECT b.ID, a.ID FROM BOOK b, AUTHOR a WHERE b.TITLE='BOOK10' AND a.NAME='AUTHOR10';
INSERT INTO BOOK_AUTHOR (BOOK_ID,AUTHOR_ID) SELECT b.ID, a.ID FROM BOOK b, AUTHOR a WHERE b.TITLE='BOOK11' AND a.NAME='AUTHOR11';
INSERT INTO BOOK_AUTHOR (BOOK_ID,AUTHOR_ID) SELECT b.ID, a.ID FROM BOOK b, AUTHOR a WHERE b.TITLE='BOOK12' AND a.NAME='AUTHOR12';
INSERT INTO BOOK_AUTHOR (BOOK_ID,AUTHOR_ID) SELECT b.ID, a.ID FROM BOOK b, AUTHOR a WHERE b.TITLE='BOOK13' AND a.NAME='AUTHOR13';
INSERT INTO BOOK_AUTHOR (BOOK_ID,AUTHOR_ID) SELECT b.ID, a.ID FROM BOOK b, AUTHOR a WHERE b.TITLE='BOOK14' AND a.NAME='AUTHOR14';
INSERT INTO BOOK_AUTHOR (BOOK_ID,AUTHOR_ID) SELECT b.ID, a.ID FROM BOOK b, AUTHOR a WHERE b.TITLE='BOOK15' AND a.NAME='AUTHOR15';
INSERT INTO BOOK_AUTHOR (BOOK_ID,AUTHOR_ID) SELECT b.ID, a.ID FROM BOOK b, AUTHOR a WHERE b.TITLE='BOOK16' AND a.NAME='AUTHOR16';
INSERT INTO BOOK_AUTHOR (BOOK_ID,AUTHOR_ID) SELECT b.ID, a.ID FROM BOOK b, AUTHOR a WHERE b.TITLE='BOOK17' AND a.NAME='AUTHOR17';
INSERT INTO BOOK_AUTHOR (BOOK_ID,AUTHOR_ID) SELECT b.ID, a.ID FROM BOOK b, AUTHOR a WHERE b.TITLE='BOOK18' AND a.NAME='AUTHOR18';
INSERT INTO BOOK_AUTHOR (BOOK_ID,AUTHOR_ID) SELECT b.ID, a.ID FROM BOOK b, AUTHOR a WHERE b.TITLE='BOOK19' AND a.NAME='AUTHOR19';
INSERT INTO BOOK_AUTHOR (BOOK_ID,AUTHOR_ID) SELECT b.ID, a.ID FROM BOOK b, AUTHOR a WHERE b.TITLE='BOOK20' AND a.NAME='AUTHOR20';
INSERT INTO BOOK_AUTHOR (BOOK_ID,AUTHOR_ID) SELECT b.ID, a.ID FROM BOOK b, AUTHOR a WHERE b.TITLE='BOOK02' AND a.NAME='AUTHOR01';
INSERT INTO BOOK_AUTHOR (BOOK_ID,AUTHOR_ID) SELECT b.ID, a.ID FROM BOOK b, AUTHOR a WHERE b.TITLE='BOOK02' AND a.NAME='AUTHOR03';
INSERT INTO BOOK_AUTHOR (BOOK_ID,AUTHOR_ID) SELECT b.ID, a.ID FROM BOOK b, AUTHOR a WHERE b.TITLE='BOOK04' AND a.NAME='AUTHOR03';
INSERT INTO BOOK_AUTHOR (BOOK_ID,AUTHOR_ID) SELECT b.ID, a.ID FROM BOOK b, AUTHOR a WHERE b.TITLE='BOOK04' AND a.NAME='AUTHOR05';
INSERT INTO BOOK_AUTHOR (BOOK_ID,AUTHOR_ID) SELECT b.ID, a.ID FROM BOOK b, AUTHOR a WHERE b.TITLE='BOOK06' AND a.NAME='AUTHOR05';
INSERT INTO BOOK_AUTHOR (BOOK_ID,AUTHOR_ID) SELECT b.ID, a.ID FROM BOOK b, AUTHOR a WHERE b.TITLE='BOOK06' AND a.NAME='AUTHOR07';
INSERT INTO BOOK_AUTHOR (BOOK_ID,AUTHOR_ID) SELECT b.ID, a.ID FROM BOOK b, AUTHOR a WHERE b.TITLE='BOOK08' AND a.NAME='AUTHOR07';
INSERT INTO BOOK_AUTHOR (BOOK_ID,AUTHOR_ID) SELECT b.ID, a.ID FROM BOOK b, AUTHOR a WHERE b.TITLE='BOOK08' AND a.NAME='AUTHOR09';
INSERT INTO BOOK_AUTHOR (BOOK_ID,AUTHOR_ID) SELECT b.ID, a.ID FROM BOOK b, AUTHOR a WHERE b.TITLE='BOOK10' AND a.NAME='AUTHOR09';
INSERT INTO BOOK_AUTHOR (BOOK_ID,AUTHOR_ID) SELECT b.ID, a.ID FROM BOOK b, AUTHOR a WHERE b.TITLE='BOOK10' AND a.NAME='AUTHOR11';
INSERT INTO BOOK_AUTHOR (BOOK_ID,AUTHOR_ID) SELECT b.ID, a.ID FROM BOOK b, AUTHOR a WHERE b.TITLE='BOOK12' AND a.NAME='AUTHOR11';
INSERT INTO BOOK_AUTHOR (BOOK_ID,AUTHOR_ID) SELECT b.ID, a.ID FROM BOOK b, AUTHOR a WHERE b.TITLE='BOOK12' AND a.NAME='AUTHOR13';
INSERT INTO BOOK_AUTHOR (BOOK_ID,AUTHOR_ID) SELECT b.ID, a.ID FROM BOOK b, AUTHOR a WHERE b.TITLE='BOOK14' AND a.NAME='AUTHOR13';
INSERT INTO BOOK_AUTHOR (BOOK_ID,AUTHOR_ID) SELECT b.ID, a.ID FROM BOOK b, AUTHOR a WHERE b.TITLE='BOOK14' AND a.NAME='AUTHOR15';
INSERT INTO BOOK_AUTHOR (BOOK_ID,AUTHOR_ID) SELECT b.ID, a.ID FROM BOOK b, AUTHOR a WHERE b.TITLE='BOOK16' AND a.NAME='AUTHOR15';
INSERT INTO BOOK_AUTHOR (BOOK_ID,AUTHOR_ID) SELECT b.ID, a.ID FROM BOOK b, AUTHOR a WHERE b.TITLE='BOOK16' AND a.NAME='AUTHOR17';
INSERT INTO BOOK_AUTHOR (BOOK_ID,AUTHOR_ID) SELECT b.ID, a.ID FROM BOOK b, AUTHOR a WHERE b.TITLE='BOOK18' AND a.NAME='AUTHOR17';
INSERT INTO BOOK_AUTHOR (BOOK_ID,AUTHOR_ID) SELECT b.ID, a.ID FROM BOOK b, AUTHOR a WHERE b.TITLE='BOOK18' AND a.NAME='AUTHOR19';
INSERT INTO BOOK_AUTHOR (BOOK_ID,AUTHOR_ID) SELECT b.ID, a.ID FROM BOOK b, AUTHOR a WHERE b.TITLE='BOOK20' AND a.NAME='AUTHOR19';
Maximum call stack size exceeded
,它指示无限循环执行(带有递归)。
最佳答案
首先,我只想说问问题的方式! :)
如您所见,“级联”部分是从父级流到子级的,而不是循环的(这可能导致堆栈溢出!抱歉,无法抗拒)。
我会给您一个解决方案,但我会先承认它不是最佳选择,因为每次更改选择列表都将需要两个Ajax调用,而不仅仅是一个。我将向APEX团队提出这个问题,希望他们将来能解决。
我将开始假设那些人已经运行了您的脚本并且有一个带有HTML区域的空白页,开始执行这些步骤。我的页面是53岁,因此人们需要相应地进行更改。
第1部分:基础知识
select title d,
id r
from book
where (
:P53_AUTHOR is null
or id in (
select book_id
from book_author
where author_id = :P53_AUTHOR
)
)
order by title
select name d,
id r
from author
where (
:P53_BOOK is null
or id in (
select author_id
from book_author
where book_id = :P53_BOOK
)
)
order by name
this.browserEvent.originalEvent !== undefined
null;
和项目,以将提交到 P53_BOOK 。该操作仅在更新操作之前用于更新P53_BOOK的 session 状态。 this.browserEvent.originalEvent !== undefined
null;
和项目,以将提交到 P53_AUTHOR 。 $('#P53_BOOK').data('last-val', $v('P53_BOOK'));
data
方法存储预刷新项的值。您将在刷新后稍后使用它。 $('#P53_AUTHOR').data('last-val', $v('P53_AUTHOR'));
$s('P53_BOOK', $('#P53_BOOK').data('last-val'), null, true);
data
方法设置了P53_BOOK的值(这次是一个 setter/getter )。传递给$s
的最后一个参数(true
)以阻止change事件,否则将创建更多循环逻辑。 $s('P53_AUTHOR', $('#P53_AUTHOR').data('last-val'), null, true);
关于javascript - APEX:如何实现循环级联选择列表(多对多),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/60676786/
我是 PHP 新手。我一直在脚本中使用 for 循环、while 循环、foreach 循环。我想知道 哪个性能更好? 选择循环的标准是什么? 当我们在另一个循环中循环时应该使用哪个? 我一直想知道要
我在高中的编程课上,我的作业是制作一个基本的小计和顶级计算器,但我在一家餐馆工作,所以制作一个只能让你在一种食物中读到。因此,我尝试让它能够接收多种食品并将它们添加到一个价格变量中。抱歉,如果某些代码
这是我正在学习的一本教科书。 var ingredients = ["eggs", "milk", "flour", "sugar", "baking soda", "baking powder",
我正在从字符串中提取数字并将其传递给函数。我想给它加 1,然后返回字符串,同时保留前导零。我可以使用 while 循环来完成此操作,但不能使用 for 循环。 for 循环只是跳过零。 var add
编辑:我已经在程序的输出中进行了编辑。 该程序要求估计给定值 mu。用户给出一个值 mu,同时还提供了四个不等于 1 的不同数字(称为 w、x、y、z)。然后,程序尝试使用 de Jaeger 公式找
我正在编写一个算法,该算法对一个整数数组从末尾到开头执行一个大循环,其中包含一个 if 条件。第一次条件为假时,循环可以终止。 因此,对于 for 循环,如果条件为假,它会继续迭代并进行简单的变量更改
现在我已经习惯了在内存非常有限的情况下进行编程,但我没有答案的一个问题是:哪个内存效率更高;- for(;;) 或 while() ?还是它们可以平等互换?如果有的话,还要对效率问题发表评论! 最佳答
这个问题已经有答案了: How do I compare strings in Java? (23 个回答) 已关闭 8 年前。 我正在尝试创建一个小程序,我可以在其中读取该程序的单词。如果单词有 6
这个问题在这里已经有了答案: python : list index out of range error while iteratively popping elements (12 个答案) 关
我正在尝试向用户请求 4 到 10 之间的整数。如果他们回答超出该范围,它将进入循环。当用户第一次正确输入数字时,它不会中断并继续执行 else 语句。如果用户在 else 语句中正确输入数字,它将正
我尝试创建一个带有嵌套 foreach 循环的列表。第一个循环是循环一些数字,第二个循环是循环日期。我想给一个日期写一个数字。所以还有另一个功能来检查它。但结果是数字多次写入日期。 Out 是这样的:
我想要做的事情是使用循环创建一个数组,然后在另一个类中调用该数组,这不会做,也可能永远不会做。解决这个问题最好的方法是什么?我已经寻找了所有解决方案,但它们无法编译。感谢您的帮助。 import ja
我尝试创建一个带有嵌套 foreach 循环的列表。第一个循环是循环一些数字,第二个循环是循环日期。我想给一个日期写一个数字。所以还有另一个功能来检查它。但结果是数字多次写入日期。 Out 是这样的:
我正在模拟一家快餐店三个多小时。这三个小时分为 18 个间隔,每个间隔 600 秒。每个间隔都会输出有关这 600 秒内发生的情况的统计信息。 我原来的结构是这样的: int i; for (i=0;
这个问题已经有答案了: IE8 for...in enumerator (3 个回答) How do I check if an object has a specific property in J
哪个对性能更好?这可能与其他编程语言不一致,所以如果它们不同,或者如果你能用你对特定语言的知识回答我的问题,请解释。 我将使用 c++ 作为示例,但我想知道它在 java、c 或任何其他主流语言中的工
这个问题不太可能帮助任何 future 的访问者;它只与一个小的地理区域、一个特定的时间点或一个非常狭窄的情况有关,这些情况并不普遍适用于互联网的全局受众。为了帮助使这个问题更广泛地适用,visit
我是 C 编程和编写代码的新手,以确定 M 测试用例的质因数分解。如果我一次只扫描一次,该功能本身就可以工作,但是当我尝试执行 M 次时却惨遭失败。 我不知道为什么 scanf() 循环有问题。 in
这个问题已经有答案了: JavaScript by reference vs. by value [duplicate] (4 个回答) 已关闭 3 年前。 我在使用 TSlint 时遇到问题,并且理
我尝试在下面的代码中添加 foreach 或 for 循环,以便为 Charts.js 创建多个数据集。这将允许我在此折线图上创建多条线。 我有一个 PHP 对象,我可以对其进行编码以稍后填充变量,但
我是一名优秀的程序员,十分优秀!