小编sah*_*har的帖子

在java中没有为jdbc找到合适的驱动程序

我是sql server的新手,想在java和sql server之间建立连接.我的连接代码是:

 public static void main(String[] args) {

    Connection con;
       try {

       String connectionUrl = "jdbc:sqlserver://HELLO-PC:1433; databaseName=Attendance Teachers;";

        con = DriverManager.getConnection(connectionUrl, "", "");
        System.out.println("connected");
        java.sql.Statement st = con.createStatement();    
      }
        catch (SQLException ex) {

        Logger.getLogger(AttendanceTeachers.class.getName()).log(Level.SEVERE, null, ex);
    }
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我的服务器名称是'HELLO-PC',我也添加了sqljdbc.jar.我看到错误:

Feb 01, 2013 11:24:46 AM attendance.teachers.AttendanceTeachers main
SEVERE: null
java.sql.SQLException: No suitable driver found for jdbc:sqlserver://HELLO-PC:1433;   databaseName=Attendance Teachers;
at java.sql.DriverManager.getConnection(DriverManager.java:604)
at java.sql.DriverManager.getConnection(DriverManager.java:221)
at attendance.teachers.AttendanceTeachers.main(AttendanceTeachers.java:30)
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我真的需要帮助.谢谢.

java sql-server connection jdbc

2
推荐指数
1
解决办法
1万
查看次数

将参数从url传递给cakephp

我需要从URL发送参数到cakephp控制器.我有两个参数'ufrom'和'uto'的消息表.在控制器中我想将此值保存在消息表中.

我输入了网址:

http://localhost/ar/messages/add?ufrom=9&uto=3
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在MessagesController我有功能:

public function add() {

if(($this->request->query['uto'])and($this->request->query['ufrom'])){
        $this->Message->create();
        if ($this->Message->save($this->request->data)) {
            $this->set('addMessage',TRUE);
            $this->set('ufrom',$this->request->query['ufrom']);
            $this->set('uto',$this->request->query['uto']);
            $this->redirect(array('action' => 'index'));
        } else {
            $this->Session->setFlash(__('The message could not be saved. Please, try again.'));
        }

        $targets = $this->Message->Target->find('list');
        $this->set(compact('targets'));
}
else{
    $this->set('error',true);
}
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}

在add.ctp我有:

<?php
if(isset($error)){
  echo('error');
}
else{
  echo json_encode($ufrom);
  echo json_encode($uto);
  echo json_encode($addMessage);
}
?>
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但是当我使用上面的URL时,我看到:

Notice (8): Undefined variable: ufrom [APP\View\Messages\add.ctp, line 6]null
Notice (8): Undefined variable: uto [APP\View\Messages\add.ctp, line 7]null
Notice (8): Undefined variable: addMessage [APP\View\Messages\add.ctp, line 8]null …
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cakephp

0
推荐指数
1
解决办法
1万
查看次数

扫到右边过渡css

我有一个提交按钮,我需要从左到右扫过它的背景:

<input type="submit" value="Send Request" class="sweep">
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我需要像扫描到正确的过渡之类的东西。所有示例都用于button 、 a 、 li 和 Div标签。

css input form-submit

0
推荐指数
1
解决办法
4497
查看次数

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cakephp ×1

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