我有一个包含项目的数组,我想做这样的事情:
<tr>
(until have items in array
<td></td><td></td>)
</tr>
但是,如果我这样做,我得到一个JSXTransformer错误:
相邻的XJS元素必须包装在一个封闭的标签中
工作版本:
{rows.map(function (rowElement){
return (<tr key={trKey++}>
<td className='info' key={td1stKey++}>{rowElement.row[0].value}</td><td key={td2ndKey++}>{rowElement.row[0].count}</td>
<td className='info' key={td1stKey++}>{rowElement.row[1].value}</td><td key={td2ndKey++}>{rowElement.row[1].count}</td>
<td className='info' key={td1stKey++}>{rowElement.row[2].value}</td><td key={td2ndKey++}>{rowElement.row[2].count}</td>
<td className='info' key={td1stKey++}>{rowElement.row[3].value}</td><td key={td2ndKey++}>{rowElement.row[3].count}</td>
<td className='info' key={td1stKey++}>{rowElement.row[4].value}</td><td key={td2ndKey++}>{rowElement.row[4].count}</td>
.......
</tr>);
})}
我试过这个.但是使用<div>封闭标签它不能正常工作.
在这里回答: 未捕获错误:不变违规:findComponentRoot(...,... $ 110):无法找到元素.这可能意味着DOM意外地发生了变异
<tbody>
{rows.map(function (rowElement){
return (<tr key={trKey++}>
{rowElement.row.map(function(ball){
console.log('trKey:'+trKey+' td1stKey'+td1stKey+' ball.value:'+ball.value+' td2ndKey:'+td2ndKey+' ball.count:'+ball.count);
return(<div key={divKey++}>
<td className='info' key={td1stKey++}>{ball.value}</td><td key={td2ndKey++}>{ball.count}</td>
</div>);
})}
</tr>);
})}
</tbody>
请告诉我如何正确包装少量TD标签! 我尝试使用指南Dynamic Children,但JSXTransformer不允许我这样做.
我在React的嵌套循环中做错了什么？我在谷歌搜索过信息,但没有找到合适的信息.你能帮我找到,我理解错了吗？
从图中可以看出,我在变量中有数据.它工作正常.但是当我添加一个不是从这个值的值时<tr>,会出现错误!
var TableBalls80 = React.createClass({
render:function(){
var rows = this.props.rows;
var columnId = 0, trKey = 0, divKey = 0, td1stKey = 0;
var td2ndKey = 100;
return(
<table className='table table-bordered bg-success'>
<thead>
<tr className='danger'>
{rows[0].row.map(function (element){
columnId++;
return (
<th colSpan="2" key={columnId}>{columnId}</th>);
})}
</tr>
</thead>
<tbody>
{rows.map(function (rowElement){
return (<tr key={trKey++}>
{rowElement.row.map(function(ball){
console.log('trKey:'+trKey+' td1stKey'+td1stKey+' ball.value:'+ball.value+' td2ndKey:'+td2ndKey+' ball.count:'+ball.count);
return(<div key={divKey++}>
<td className='info' key={td1stKey++}>{ball.value}</td><td key={td2ndKey++}>{ball.count}</td>
</div>);
})}
</tr>);
})}
</tbody>
</table>);
}
});
错误(取决于从另一个项目添加的项目<tr>):
未捕获错误:不变违规:findComponentRoot(...,.0.1.1.0.2.0.0.1.$ 0. …
我有直接的List list1. List<String> list = Ordering.natural().sortedCopy(asu2);
如何改变秩序.我不知道如何从extends类重写方法,请用例子写或清楚说明.
我想知道为什么我的代码运行方式与我预期的有点不同.我按下按钮并调试:
Disconnected to the target VM, address: 127.0.0.1:64040 transport: socket
IDE:IntelliJ Idea 12,操作系统:Windows 7
cmd>netstat
TCP 127.0.0.1:62522 T145:62523 ESTABLISHED
TCP 127.0.0.1:62523 T145:62522 ESTABLISHED
TCP 127.0.0.1:63544 T145:nfsd-status ESTABLISHED
TCP 127.0.0.1:65084 T145:nfsd-status ESTABLISHED
TCP 127.0.0.1:65086 T145:nfsd-status ESTABLISHED
TCP 127.0.0.1:65458 T145:nfsd-status ESTABLISHED
cmd>ping 127.0.0.1:64040 - timeout
请给我建议如何正确设置Jetty.我正在使用最新版本的jetty(9.0.6.v20130930).我订购了服务器实现具体的servlet,它不起作用!我已经和Jetty做了一点工作,没有看到类似的东西.
- Main.class
public static void main(String[] args) throws Exception
{
MessageSystem ms = new MessageSystem();
Frontend frontend = new Frontend(ms);
(new Thread(frontend)).start();
Server server = new Server(8080);
server.setHandler(frontend);
server.start();
server.join();
}
- Frontend.class
public class Frontend extends AbstractHandler implements Runnable, Abonent {
... constructor and others methods...
public void handle(String target, Request baseRequest,
HttpServletRequest request,
HttpServletResponse response)
throws IOException, ServletException
{
setResponseSettings(baseRequest,response);
PrintWriter out = response.getWriter();
try {
int id;
HttpSession session = request.getSession();
if( session.isNew()){
id = setSessionId(session); …给定一个可能包含重复项的列表(如下所示),我需要能够计算每个(关键字)数量的唯一元素.
List<String> list = new ArrayList<String>();
Set<String> set = new HashSet<String>();
list.add("M1");
list.add("M1");
list.add("M2");
list.add("M3");
set.addAll(list);
System.out.println(set.size());
如何从列表中获取每个唯一元素的计数？这意味着我想知道列表(列表)中包含多少"M1","M2"等多少.
The result should be the following:
2 M1
1 M2
1 M3
case class Account(var email:String, var pass:String, var familyId: Int, var accessId: Int, id: Option[Int] = None)
// A Accounts table with 5 columns: id, email, pass, familyId, accessId
class Accounts(tag: Tag) extends Table[Account](tag, "ACCOUNTS") {
def id = column[Int]("ID", O.AutoInc, O.PrimaryKey)
def email = column[String]("EMAIL")
def pass = column[String]("PASS")
def familyId = column[Int]("FAMILY_ID") // TODO: add fk family_id
def accessId = column[Int]("ACCESS_ID") // TODO: add fk access_id
def * = (email, pass, familyId, accessId, id.?) <> (Account.tupled, Account.unapply)
}
创建具有这种结构的表后,表中的所有列都是 …
例如:String[] str = {"M1","M1","M1","M2","M3"};
最推荐的是答案 - HashSet.哪种方法还是你有更好的想法？