我正在做一些HTML和JQuery的工作.我有一个问题,我的textarea和选择单选按钮后没有出现提交按钮.我的HTML看起来像这样:
<html>
<head><title>Publications Database | Which spotlight for Publications</title>
<script type="text/javascript" src="./jquery.js"></script>
<script src="./addSpotlight.js" type="text/javascript" charset="utf-8"></script>
</head>
<body>
<div class="wrapper">
<div class="header">
<div class="headerText">Which spotlight for Publications</div>
</div>
<div class="mainContent">
<p>Please select the publication that you would like to make the spotlight of this month:</p>
<form action="addSpotlight" method="POST" id="form" name="form">
<div class="div29" id="div29"><input type="radio" value="29" name="publicationIDs" >A System For Dynamic Server Allocation in Application Server Clusters, IEEE International Symposium on Parallel and Distributed Processsing with Applications, 2008 </div>
<div class="div30" …我希望能够检查我的结果集中是否有结果.Todo这将我执行:
if(rs.next()){
boolean = true;
}
但是,我想检查是否值在数据库中,如果是这样检索它:
while(rs.next())
id = rs.getInt("id);
我如何将这两者结合起来？如果结果集为空,我希望返回-2.
迪恩
,先谢谢你
本周早些时候我问了一个问题,我得到了这个陈述作为答案:
select published, count(*) nbr
from table1
group by published
order by nbr desc
limit 1
我现在想知道如何在语句中添加where子句,以便限制我得到的不同类型的出版物的结果.我的表看起来像这样:
where子句将在type列上,例如where type=3.
先谢谢
Dean
我有一个类型模块的ArrayList,每个模块都有一个分配的arraylist.我写了下面的TableModel,但是当在表中选择了某些东西时,我遇到了问题,导致我的ArrayLists超出范围.这是我的桌子型号:
public class AssignmentsTableModel extends AbstractTableModel {
private ArrayList<Module> modules;
private static final String[] COLUMN_NAMES = {"Module Identifier", "Module Name", "Assignment Title", "Author", "Date Set", "Date Due", "Weighting"};
private int moduleID;
private int assignmentID;
private Module module;
private int totalNumberAssignments;
public AssignmentsTableModel(ArrayList<Module> modules) {
moduleID = 0;
assignmentID = 0;
this.modules = modules;
module = modules.get(moduleID);
totalNumberAssignments = getRowCount();
}
public AssignmentsTableModel(Module module) {
modules = new ArrayList<Module>();
modules.add(module);
}
@Override
public int getRowCount() {
int rowCount = 0;
for(Module …我正在尝试将文本写入字节但是当我打电话时:
Charset charset = Charset.forName("UTF-16");
ByteBuffer bytes = charset.encode(toConvert.trim());
byte[] charsAsBytes = bytes.array();
我得到一个以-2和-1开头的数组.这些价值来自哪里？他们是什么意思,或者我可以跳过这些.默认字符集是UTF-8.
我有一个16位进入USB的无符号整数,这些整数在我的列表中显示为2个字节,例如[0x20, 0x00].我如何将这些组合到python中的数字？
我一直收到错误:
_tkinter.TclError:未知选项"-menu"
我的MWE看起来像:
from tkinter import *
def hello():
print("hello!")
class Application(Frame):
def createWidgets(self):
self.menuBar = Menu(master=self)
self.filemenu = Menu(self.menuBar, tearoff=0)
self.filemenu.add_command(label="Hello!", command=hello)
self.filemenu.add_command(label="Quit!", command=self.quit)
def __init__(self, master):
Frame.__init__(self, master)
self.pack()
self.createWidgets()
self.config(menu=self.menuBar)
if __name__ == "__main__":
root = Tk()
ui = Application(root)
ui.mainloop()
我在OS X 10.8上使用python 3.为什么我收到未知选项错误？
java ×3
python ×2
dom ×1
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