我无法让tomcat7编译jsps.直到运行示例servlet就好了,服务启动并运行.我正在运行oracle java 8.
谁能指出我正确的方向？
这是堆栈跟踪:
type Exception report
message Unable to compile class for JSP:
description The server encountered an internal error that prevented it from fulfilling this request.
exception
org.apache.jasper.JasperException: Unable to compile class for JSP:
An error occurred at line: 1 in the generated java file
The type java.util.Map$Entry cannot be resolved. It is indirectly referenced from required .class files
Stacktrace:
org.apache.jasper.compiler.DefaultErrorHandler.javacError(DefaultErrorHandler.java:102)
org.apache.jasper.compiler.ErrorDispatcher.javacError(ErrorDispatcher.java:331)
org.apache.jasper.compiler.JDTCompiler.generateClass(JDTCompiler.java:468)
org.apache.jasper.compiler.Compiler.compile(Compiler.java:378)
org.apache.jasper.compiler.Compiler.compile(Compiler.java:353)
org.apache.jasper.compiler.Compiler.compile(Compiler.java:340)
org.apache.jasper.JspCompilationContext.compile(JspCompilationContext.java:646)
org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:357)
org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:390)
org.apache.jasper.servlet.JspServlet.service(JspServlet.java:334)
javax.servlet.http.HttpServlet.service(HttpServlet.java:728)
note The full stack trace of …我试图添加一列到从中创建的数组recfromcsv.在这种情况下,它是一个数组:( [210,8]rows,cols).
我想添加第九列.空或用零并不重要.
from numpy import genfromtxt
from numpy import recfromcsv
import numpy as np
import time
if __name__ == '__main__':
print("testing")
my_data = recfromcsv('LIAB.ST.csv', delimiter='\t')
array_size = my_data.size
#my_data = np.append(my_data[:array_size],my_data[9:],0)
new_col = np.sum(x,1).reshape((x.shape[0],1))
np.append(x,new_col,1)
我无法使用XML序列化将动物列表保存到磁盘.
我得到Exception:Thrown:"类型AnimalLibrary.Animals.Mammals.Dog不是预期的.使用XmlInclude或SoapInclude属性来指定静态未知的类型." (System.InvalidOperationException)
如果我使用"Dog"尝试注释代码,它将按预期工作,并生成XML.但是作为列表中唯一元素发送的同一只狗不起作用.
[XmlElement("animalList")]
public List<Animal> animalList = new List<Animal>();
public bool SaveBinary(string fileName)
{
Mammals.Dog dog = (Mammals.Dog)animalList[0];
//IObjectSerializer<Mammals.Dog> obj = new XMLObjectSerializer<Mammals.Dog>();
IObjectSerializer<List<Animal>> obj = new XMLObjectSerializer<List<Animal>>();
bool saved = obj.SaveFile(fileName, animalList);
if (saved)
return true;
return false;
}
XML序列化器
public bool SaveFile(string fileName, T objectToSerialize)
{
try
{
//Will overwrite old file
XmlSerializer mySerializer = new XmlSerializer(typeof(T));
StreamWriter myWriter = new StreamWriter(fileName);
mySerializer.Serialize(myWriter, objectToSerialize);
myWriter.Close();
}
catch (IOException ex)
{
Console.WriteLine("IO Exception ", ex.Message);
return false; …我似乎可以获得对数据库的jpa查询的语法正确.使用squirrel SQL可以很好地工作.
数据库是Derby,代码使用的是JPA.
更新了新的查询和错误.这让我相信它在实体映射中是不正确的.它可能是joinColumn子句中的某些内容？
修正了命名.这导致了第一个问题,即没有使用正确的实体名称.
Query q = em.createQuery("select t, sum(t.result) from Serie t, Player p " +
" where p.id = t.player" +
" group by t.player");
Exception [EclipseLink-6076] (Eclipse Persistence Services - 2.5.0.v20130507-3faac2b): org.eclipse.persistence.exceptions.QueryException
Exception Description: Object comparisons can only be used with OneToOneMappings. Other mapping comparisons must be done through query keys or direct attribute level comparisons.
Mapping: [org.eclipse.persistence.mappings.DirectToFieldMapping[id-->PLAYER.ID]]
Expression: [
Query Key id
Base com.jk.hcp.Player]
Query: ReportQuery(referenceClass=Serie jpql="select t, sum(t.result) from Serie t, Player p where …我试图想出一种在单个SQL中执行以下操作的方法.
我想在09:00到17:30之间以1分钟为增量选择所有时间而不使用for循环.有办法吗？
select avg(price), Date(timecolumn)
from table
where Time(timeColumn) >= '09:00:00'
and Time(timeColumn) <= '09:01:00'
我正在尝试使用 docker-compose 从 docker 内部连接到rabbitMQ。从容器外部运行的程序连接到“image:rabbitmq:3-management”没有问题。
没有设置密码或任何东西。
仔细查看错误消息后，鼠兔似乎尝试使用 IPv6 而不是 IPv4 连接。查看文档我无法找到连接 IPv4 的方法。
故障发生于
self.connection = pika.BlockingConnection(pika.ConnectionParameters(host='localhost'))
Error log
rabbitMQ_1 | 2020-12-09 12:41:42.332 [info] <0.685.0> Ready to start client connection listeners
rabbitMQ_1 | 2020-12-09 12:41:42.333 [info] <0.988.0> started TCP listener on [::]:5672
listener_1 | ERROR:pika.adapters.utils.io_services_utils:Socket failed to connect: <socket.socket fd=9, family=AddressFamily.AF_INET, type=SocketKind.SOCK_STREAM, proto=6, laddr=('127.0.0.1', 36782)>; error=111 (Connection refused)
listener_1 | ERROR:pika.adapters.utils.connection_workflow:TCP Connection attempt failed: ConnectionRefusedError(111, 'Connection refused'); dest=(<AddressFamily.AF_INET: 2>, <SocketKind.SOCK_STREAM: 1>, 6, '', ('127.0.0.1', 5672))
listener_1 | ERROR:pika.adapters.utils.connection_workflow:AMQPConnector - reporting …我似乎无法让tomcat编译它应该服务的jsp文件.它将完美地运行.war文件和servlet.但是在任何jsp页面都会失败.我得到的印象是它没有找到常规的java包含.
我正在尝试使用此文件(或包含的任何其他jsp测试文件)
<%@ page language="java" import="java.util.*" errorPage="" %>
<html>
<body>
Current Date time: <%=new java.util.Date()%>
</body>
</html>
type Exception report
message Unable to compile class for JSP:
description The server encountered an internal error that prevented it from fulfilling this request.
exception
org.apache.jasper.JasperException: Unable to compile class for JSP:
An error occurred at line: 1 in the generated java file
The type java.util.Map$Entry cannot be resolved. It is indirectly referenced from required .class files
Stacktrace:
org.apache.jasper.compiler.DefaultErrorHandler.javacError(DefaultErrorHandler.java:102)
org.apache.jasper.compiler.ErrorDispatcher.javacError(ErrorDispatcher.java:331)
org.apache.jasper.compiler.JDTCompiler.generateClass(JDTCompiler.java:468)
org.apache.jasper.compiler.Compiler.compile(Compiler.java:378)
org.apache.jasper.compiler.Compiler.compile(Compiler.java:353) …我正在使用“for of”循环。但它仍然没有等待虚拟函数退出。我假设等待会等待“虚拟”函数完成，然后使用“for of”。
日志输出：
End wait true
starting wait
End wait true
starting wait
End wait true
I am waiting for 925.6301720227887
I am waiting for 923.6969211579702
I am waiting for 962.0987671698102
etc...
const dummy = async(timeToWait) => {
await setTimeout(() => {
console.log("I am waiting for", timeToWait);
}, timeToWait);
return Promise.resolve(true);
}
// Iterate over the minutes off and get aggregate data
const computeAggregate = async (model, sym) => {
await model.find({"symbol": sym})
.sort({trade_ts:1}).exec()
.then(async (symbol) => {
var …我正在尝试对地图进行排序以显示下拉列表.但我无法完成任何排序.这将返回一张新地图.但不是像我期望的那样用按键排序的地图.
private Map<String, String> mapInstrumentIDs = new TreeMap<>();
Map<Object, Object> val = mapInstrumentIDs
.entrySet()
.stream()
.sorted(Map.Entry.comparingByKey())
.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));
我当然没有想到密钥实际上是一个整数.这意味着将其排序为字符串不会给我预期的结果(作为整数排序).将键更改为Integer并转换该值将产生预期结果.
问题是为什么变量测试为空？该程序将返回正确排序的数组.但它似乎没有被分配.
def my_sort(array_to_sort):
sort = False
number_of_items = len(array_to_sort)
print "sorted array: ", array_to_sort
for i in range(0, number_of_items-1):
if array_to_sort[i] > array_to_sort[i+1]:
tmp = array_to_sort[i]
array_to_sort[i] = array_to_sort[i+1]
array_to_sort[i+1] = tmp
sort = True
if sort == True:
my_sort(array_to_sort)
elif sort == False:
return array_to_sort
if __name__ == '__main__':
# main()
arr = [4,5,7,3,2,1]
test = my_sort(arr)
print (test)
这将返回以下内容.
sorted array: [4, 5, 7, 3, 2, 1]
sorted array: [4, 5, 3, 2, 1, 7]
sorted array: [4, …我想输出一个primefaces数据表.
<p:dataTable var="bi" value="#{ bowlingInfo.hallMastaren} ">
<p:column headerText="Position">
<h:outputText value="#{bi.position }" />
</p:column>
</p:dataTable>
但它不会识别它看起来的数据,并抛出以下异常.
javax.servlet.ServletException:/test.xhtml @ 13,47 value ="#{bi.position}":在类型java.lang.String上找不到属性'position'
我正在使用primefaces 5.1.
删除<h:outputText value="#{bi.position }" />将呈现页面.但它只会渲染一次行.由于我添加了两条记录,所以它期望它呈现两行.