小编Kev*_*vin的帖子

        $query = "SELECT *
              FROM users
              WHERE username = '".mysql_escape_string($username)."'
              AND password = '".mysql_escape_string($password)."'";
        $result = mysql_fetch_array(mysql_query($query));

现在这是我对一般访问者的查询,现在如果我需要管理员和主持人,我如何从具有特定ID的相同用户处理.

我的SQL查询.

CREATE TABLE IF NOT EXISTS `users` (
  `yourID` int(10) unsigned NOT NULL auto_increment,
  `username` varchar(26) collate latin1_general_ci NOT NULL,
  `password` varchar(26) collate latin1_general_ci NOT NULL,
   PRIMARY KEY  (`yourID`)
) ENGINE=MyISAM  DEFAULT CHARSET=latin1 COLLATE=latin1_general_ci AUTO_INCREMENT=87992 ;

管理员和管理员的登录页面将相同.

$result = mysql_query("SELECT * FROM project ORDER BY projectid");

while($row = mysql_fetch_array($result)) 
{
    return(array($row['projectid'],   $row['clientname'], 
                 $row['salesperson'], $row['prospect']));       
}

我只从字段中获得第一组值.我需要所有的价值观.

如何使用PHP从LDAP获取所有用户.

function getUsers() {

         $process = new process();
    //ldap->s_Host('domain.com');
    //$ldap->s_Domain('domain.com');
        $process->s_Host('domain.com');
        $process->s_Domain('domain.com');   
        $process->s_LdapSecure(false);

         if($process->LdapConn()){
             if($process->LdapBind()){



     }

}
}

我有一个需要传递的项目列表,应该删除.我怎么能用PHP做到这一点？

$query = "DELETE from members WHERE member_id ='$chan[2]' ";

$chan[2] 包含多个值,但只删除一个值.

以下SQL查询有什么问题？

$sql="SELECT * FROM ".TABLE." WHERE desgid='$id', weightid='$weightid' AND deptid='$deptid' ";

错误消息是:

您的SQL语法有错误; 查看与您的MySQL服务器版本对应的手册,以便在第1行'weightid ='12'和deptid = '31''附近使用正确的语法

我有一个XML,它有一个属性选项或组合框,解析我需要动态地在我的flex中创建组件.

Viatropos给出了一个很棒的代码,但是我无法执行它...任何人都可以生成它...谢谢

@Override
public void onCreateOptionsMenu(Menu menu, MenuInflater inflater) {
    Log.d("Does", "get called");
    inflater.inflate(R.menu.menuitem, menu);
    super.onCreateOptionsMenu(menu, inflater);
}

下面是我的onCreateView方法,我正在调用

@Override
public View onCreateView(LayoutInflater inflater, 
        ViewGroup container, Bundle savedInstanceState) {   
          setHasOptionsMenu(true);
          return inflater.inflate(R.layout.layout1, container, false);
}

我没有在我的操作栏中获取日志语句或菜单.

更新:我尝试从片段的onCreate方法调用它,但菜单未显示.

@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setHasOptionsMenu(true);        
}

menu.xml文件

<menu xmlns:android="http://schemas.android.com/apk/res/android">

    <item android:id="@+id/section" android:title="@string/section"
        android:icon="@drawable/ic_section"
        android:showAsAction="always" />

    <item android:id="@+id/refresh" android:title="@string/refresh" 
        android:icon="@drawable/ic_refresh"
        android:showAsAction="always" />

    <item android:id="@+id/edit_patient" android:title="@string/edit_patient" 
        android:icon="@drawable/ic_editpatient"
        android:showAsAction="always" />    

    <item android:id="@+id/about" android:title="@string/about"
        android:showAsAction="never" />

    <item android:id="@+id/help" android:title="@string/help"
        android:showAsAction="never" />

    <item android:id="@+id/signout" android:title="@string/signout"
        android:showAsAction="never" …

HttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost(URL);

try {
    StringEntity documentStringified = new StringEntity(Obj.toString());
    httpPost.setEntity(documentStringified);
} catch (UnsupportedEncodingException e) {
    Log.d("UnsupportedEncodingException", e.toString());
}

try {
    HttpResponse response = httpClient.execute(httpPost);
    Log.d("Response", response.toString());
} catch (IOException e) {
    Log.d("IOException", e.toString());
}

我无法得到response.如何打印响应Logger or console. response.toString()或response.getEntity.toString()不起作用.

我应该将Content-type设置为"application/json".

https://github.com/jfeinstein10/SlidingMenu

我正在使用这个库,我想在我打开一个新的时隐藏滑动菜单activity,这样当我按回来时我不希望滑动菜单出现.

@Override
public void onListItemClick(ListView lv, View v, int position, long id) {
    switch (position) {
    case 0:
        Intent intent = new Intent("android.intent.action.Home");
        getActivity().startActivity(intent);
        break;
   }
}

上面的代码在我的SherlockListFragment班级,我想隐藏滑动菜单,一旦我开始一个新的活动.

$result = mysql_query("INSERT INTO project (clientname, salesperson, prospect)
                       VALUES ('$clientName','$salesPer','$prospectVal')");

            while ($row = mysql_fetch_assoc($result)) {

                $projectID = $row['projectid'];


            return $projectID;

        }

我没有从数据库中获取projectID.字段名称是动态名称,在每次插入时递增.