我在tableView中有这个自定义的CellFactory.滚动时,该列异常缓慢.任何理由都是这样,我该如何改进它.
lastTradeColumn.setCellFactory(
new Callback<TableColumn<Stock, Price>,TableCell<Stock, Price>>(){
@Override public TableCell<Stock, Price> call( TableColumn<Stock, Price> p ) {
TableCell<Stock, Price> cell = new TableCell<Stock, Price>() {
@Override public void updateItem(Price price, boolean empty) {
super.updateItem(price, empty);
if (price != null) {
VBox vbox = new VBox(5);
vbox.getChildren().add(new Label("£"+price.toString()));
if( price.getOldPrice() > price.getNewPrice()) {
vbox.setStyle("-fx-background-color:#EA2A15;");
}
else if( price.getOldPrice() < price.getNewPrice()) {
vbox.setStyle("-fx-background-color:#9CF311;");
}
setGraphic( vbox );
}
}
};
return cell;
}
});
Run Code Online (Sandbox Code Playgroud) 我似乎无法弄清楚为什么它打印不正确:
public class test3 {
public static void main(String[] args) {
double[] gens = {100, 200.1, 9.3, 10};
double d0 = 0;
double d1 = 0;
double d2 = 0;
double d3 = 0;
double d4 = 0;
double d5 = 0;
double d6 = 0;
double d7 = 0;
double d8 = 0;
double d9 = 0;
for (int i = 0; i < gens.length; i++) {
double percs = gens[i];
while (percs < -9 || 9 < percs) …Run Code Online (Sandbox Code Playgroud) 我正在尝试发送查询网址
String url = String.format(
"http://xxxxx/xxx/xxx&message=%s",myEditBox.getText.toString());
// Create a new HttpClient and Post Header
DefaultHttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost(url);
httpclient.getCookieStore().addCookie(cooki);
try {
ResponseHandler<String> responseHandler = new BasicResponseHandler();
httpclient.getParams().setParameter("http.connection-manager.timeout", 15000);
String response = httpclient.execute(httppost, responseHandler);
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给我错误,illegal character at query.那white space可能是.如何处理这个问题?
最好的祝福
嗨!我有这个问题:我必须在java中创建一个程序设计一个人的图形,我必须绘制它.我已经编写了设计人类的代码,但我不知道如何用颜色填充形状.我知道我必须使用"java.awt.Color",但我不知道如何.
颜色必须是:图像背景(黄色),头部(蓝色),手臂和腿部(绿色),身体(红色).
到目前为止,这是我的代码:
import javax.swing.*;
import java.awt.*;
public class DrawPanelTest {
//creates a window to display the drawing
public static void main(String[] args) {
// create a new frame to hold the panel
JFrame application = new JFrame();
Container pane=application.getContentPane();
// create a panel that contains our drawing
DrawPanel panel = new DrawPanel();
// set the frame to exit when it is closed
application.setDefaultCloseOperation( JFrame.EXIT_ON_CLOSE );
// add the panel to the frame
pane.add(panel);
application.setContentPane(pane);
// set the size of …Run Code Online (Sandbox Code Playgroud) 我要插入的参数数量不定,(2031) No data supplied for parameters in prepared statement在警告后出现错误Number of variables doesn't match number of parameters in prepared statement in SaveIntermediateData.php5 on line 49。
$link = new mysqli( DB_HOST, DB_USER, DB_PASSWORD, DB_NAME );
if( ! $link ) {
echo "<h1>new mysqli() failed!</h1>";
exit( 0 );
}
$queryText =
"CREATE TABLE IF NOT EXISTS visitors (".
"id VARCHAR( 512) CHARACTER SET ASCII NOT NULL,".
"name VARCHAR( 80) CHARACTER SET ASCII NOT NULL,".
"value VARCHAR(4096) NOT NULL,".
"PRIMARY …Run Code Online (Sandbox Code Playgroud) 使用案例:
启动时列表:

添加18后:

更新后:

你可以看到没有任何变化!
代码:
public final class SortedListTest extends Application {
@Override
public void start( Stage stage ) throws Exception {
final ObservableList<IntegerProperty> il =
FXCollections.observableArrayList();
il.add( new SimpleIntegerProperty( 12 ));
il.add( new SimpleIntegerProperty( 24 ));
il.add( new SimpleIntegerProperty( 36 ));
final Button add = new Button( "Add 18" );
final Button update = new Button( "Update 24 to 8" );
final HBox ctrl = new HBox( 4.0, add, update );
final ListView<IntegerProperty> listVw = …Run Code Online (Sandbox Code Playgroud) 我现在正在学习HTML/CSS,所以我做了一点点cookie-clicker看起来很有趣:http://jsfiddle.net/Zerh/6kb87sp9/
我的问题是:如果我在任何按钮上点击太快(双击),它会突出显示一个选项.我使用以下CSS来隐藏突出显示:
::selection {
background: transparent;
}
::-moz-selection {
background: transparent;
}
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它似乎适用于Firefox/Edge,但我不能让它在Chrome上运行.
我发现了这个老话题:如何使用CSS禁用文本选择突出显示?,但它已经6岁了,似乎对我不起作用.
这是来源:
<!doctype html>
<html>
<head>
<meta charset="UTF-8" />
<title>White</title>
<style>
body {
background-color: black;
}
.quadrant-legend {
position: absolute;
left : 28px;
width : 288px;
height : 62px;
}
.quadrant-legend-img {
position: absolute;
top : 0;
left : 0;
z-index : 3;
}
.quadrant-legend-btn {
position: absolute;
top : 6px;
right : 34px;
z-index : 3;
}
.quadrant-legend-text {
margin-left: 16px;
margin-top : 16px;
font-family: arial, sans-serif;
font-size : 26px;
color : white;
z-index : 4;
}
#quadrant-legend-asset {
top: …Run Code Online (Sandbox Code Playgroud) 我的代码使用 GCC 生成语法错误:
Run Code Online (Sandbox Code Playgroud)src/main.cpp: In function ‘int main()’: src/main.cpp:95:4: error: could not convert ‘{{"enum", E_PRINT}, {"string", "setup"}, {"object", {{"double", 3.1415926535897931e+0}, {"long", 1235813l}}}}’ from ‘<brace-enclosed initializer list>’ to ‘Object’ };
#include <cmath>
#include <initializer_list>
#include <iostream>
#include <memory>
#include <sstream>
#include <string>
#include <vector>
#include <stdlib.h>
struct NamedValueBase {
NamedValueBase( const std::string & name ) :
name( name )
{}
virtual ~ NamedValueBase( void ) {}
std::string name;
};
template<class T>
struct NamedValue : public NamedValueBase {
NamedValue( const std::string & …Run Code Online (Sandbox Code Playgroud) 请参阅java.util.Collections.reverse()的文档.
我不明白Java方法如何只能通过值传递,但是允许以下内容:
public static void reverse(List<?> list)
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它如何改变通过参数传递的原始对象?特别是当通过参数传递的所有对象都按值传递时?这不应该意味着只传递对象的副本,因此只能返回一个引用吗?由于我们返回无效,我们如何更改传递的原始对象?