我正在尝试从数据库中选择记录，其中记录 ID 不在列表中。
看看下面@我的问题。
String Sqlids = "2,6,3,9"; // this is dynamic so the number of element is unknown
String str= "SELECT TOP 1 * FROM student WHERE ID NOT IN (2,6,3,9) ORDER BY NEWID()";
PreparedStatement stat = con.prepareStatement(str);
ResultSet rs = stat.executeQuery();
上面的语句工作正常，但如果我将其更改为
String Sqlids = "2,6,3,9";
String str= "SELECT TOP 1 * FROM student WHERE ID NOT IN (Sqlids) ORDER BY NEWID()";
PreparedStatement stat = con.prepareStatement(str);
ResultSet rs = stat.executeQuery();
//i also try this
String Sqlids = "2,6,3,9"; …触发后，管道运行并返回以下错误，
$ ssh-add <(echo "$SSH_PRIVATE_KEY")
Enter passphrase for /dev/fd/63: ERROR: Job failed: exit code 1
我的SSH_PRIVATE_KEY变量具有有效的私钥。
我有一个扩展jlabel并使用paintComponent绘制它的类,如下所示:paintPhotos.java
package myApp;
import java.awt.*;
import javax.swing.*;
/**
*
* @author PAGOLINA
*/
public class paintPhotos extends javax.swing.JLabel {
public Image img; int w; int h;
public paintPhotos(Image img, int w, int h) {
this.img = img; this.w = w; this.h = h;
System.out.println("am paintclass");
}
@Override
public void paintComponent(Graphics p) {
System.out.println("am here");
super.paintComponent(p);
Graphics2D g2 = (Graphics2D) p;
p.drawImage(img, 0, 0, w, h, this);
}
}
当我尝试从另一个类的构造函数中绘制时(AddScore.java).
public AddScore() {
initComponents();
setLocationRelativeTo(null);
removeNotify();
setUndecorated(true);
Image imag = new …