小编Joh*_*van的帖子

我有一些ID为"div1","div2"......"divx"的DIV.我想对它们进行操作.有没有办法获得元素ID .contains"div".还有其他想法吗？

我有一个表有一个类型varchar的coulmn我需要将它转换为Nvarchar但它显示一个错误,我必须删除表并重新创建它:)任何建议!

我在GWT中有一个方法,它使用请求的fire方法从数据库中检索DATA,因为你们都知道它的异步我从JS调用这个方法所以我需要使同步成为可能

private static String retriveLocation(String part)
{
    ClientFactory clientFactory = GWT.create(ClientFactory.class);
    MyRequestFactory requestFactory = clientFactory.getRequestFactory();
    YadgetRequest request = requestFactory.yadgetRequest();
    String criteria = "!" + part;
    final ArrayList<String> tags = new ArrayList<String>();

    request.getTagsStartingWith(criteria, 10, 0).fire(
            new Receiver<List<TagProxy>>() {
                @Override
                public void onSuccess(List<TagProxy> tagList) {
                    String output = "[";

                    for (TagProxy pt : tagList) {
                        output += "{";
                        output += "\"id\":" + "\"" + pt.getId() + "\",";
                        output += "\"value\":"
                                + "\""
                                + pt.getName().replaceAll("\"", "")
                                        .replaceAll("!", "") + "\"";
                        output += "},"; …

我正在开发一个原生的Android应用程序,每30分钟运行一次操作.我需要确保处理器正在运行,所以我使用了

警报接收器内的PowerManager.Wakelock.

这是我在警报接收器里面使用的代码.

public class AlarmReceiver extends BroadcastReceiver {
    @Override
    public void onReceive(Context context, Intent arg1) {

        PowerManager pm = (PowerManager) context.getSystemService(context.POWER_SERVICE);
        PowerManager.WakeLock wl = pm.newWakeLock(PowerManager.PARTIAL_WAKE_LOCK, "My Tag");
        wl.acquire();
        Intent eventService = new Intent(context, SyncInBackground.class);
        context.startService(eventService);
        wl.release();
    }
}

在我的主AndroidManifest.xml中

我添加了以下行来请求权限

<uses-permission android:name="android.permission.wake_lock"/>

当我在我的应用程序中运行警报时,我得到以下异常

02-07 23:25:02.839  13699-13699/com.snappext E/AndroidRuntime? FATAL EXCEPTION: main
    java.lang.RuntimeException: Unable to start receiver com.snappext.Backup.AlarmReceiver: java.lang.SecurityException: Neither user 10008 nor current process has android.permission.WAKE_LOCK.
            at android.app.ActivityThread.handleReceiver(ActivityThread.java:2140)
            at android.app.ActivityThread.access$1500(ActivityThread.java:127)
            at android.app.ActivityThread$H.handleMessage(ActivityThread.java:1211)
            at android.os.Handler.dispatchMessage(Handler.java:99)
            at android.os.Looper.loop(Looper.java:137)
            at android.app.ActivityThread.main(ActivityThread.java:4512)
            at java.lang.reflect.Method.invokeNative(Native …

我有一个非常简单的 JSF 2.2 ( hibernate 4.3 ) 应用程序，它有 2 个页面。第一页是 login.xhtml，它绑定到会话 bean 下面

import javax.inject.Named;
import java.io.Serializable;
import java.util.logging.Level;
import java.util.logging.Logger;
import javax.enterprise.context.SessionScoped;
import javax.faces.application.FacesMessage;

import javax.faces.context.FacesContext;


@Named(value = "loginBean")
@SessionScoped
public class LoginBean implements Serializable {

    /**
     * Creates a new instance of LoginBean
     */
    Company company;
   
    private static final long serialVersionUID = 1520318172495977648L;

    public String Login() {
        CompanyHelper companyHelper = new CompanyHelper();
        company = companyHelper.Login(this.loginName, this.password);
        if (company != null) {
            return "";
        } else {
            FacesMessage msg …

我没有任何 python 编程经验，但我可以运行 python 文件，有没有人可以帮助我，这里是文件内容

>>>>>>>>>>>>>>
#!/usr/local/bin/python

"""
Based on: http://wxpsvg.googlecode.com/svn/trunk/svg/pathdata.py
According to that project, this file is licensed under the LGPL
"""
try:
    from pyparsing import (ParserElement, Literal, Word, CaselessLiteral, 
        Optional, Combine, Forward, ZeroOrMore, nums, oneOf, Group, ParseException, OneOrMore)
except ImportError:
    import sys
    sys.exit("pyparsing is required")


#ParserElement.enablePackrat()

def Command(char):
    """ Case insensitive but case preserving"""
    return CaselessPreservingLiteral(char)

def Arguments(token):
    return Group(token)


class CaselessPreservingLiteral(CaselessLiteral):
    """ Like CaselessLiteral, but returns the match as found
        instead of as defined.
    """
    def __init__( …

我有一个带有一个字段的表单,emai有一个非常简单的验证required = true.现在,当我点击按钮时,它会执行AJAX回调.我想知道无论如何在客户端进行验证而不调用服务器？

<h:form class="lfrm" id="lfrm">
    <p:panel id="lpanel" header="Login">
        <p:messages id="lmessages" showDetail="true" autoUpdate="true"/>

        <h:panelGrid id="lgrid" columns="2" cellpadding="5">
            <h:outputText value="Email: *" />

            <p:inputText styleClass="email"
                         id="lemail"
                         value="#{Registration.user.email}"
                         required="true" requiredMessage="Email is required">
            </p:inputText>

            <p:commandButton process="lemail"
                             value="Sign in"
                             immediate="true"
                             validateClient="true"/>
    </p:panel>
</h:form>