小编soo*_*man的帖子

有人可以告诉我一个如何使用 criteriaBuilder 进行 Hibernate 查询并连接到子查询的示例吗

这是一个简单的例子：

select idSchool, name from user left join (select idSchool from student where age < 15) as NewTable on idSchool = idSchool

谢谢

是否可以@Value从另一个变量设置

例如.

System properties : firstvariable=hello ,secondvariable=world

@Value(#{systemProperties['firstvariable'])
String var1;

现在我想要var2与var1它连接并依赖它,就像这样

    @Value( var1 + #{systemProperties['secondvariable']
    String var2;

public void message(){ System.out.printlng("Message : " + var2 );

我FormLayout和CheckBox组件有问题.我想用它FormLayout来为用户生成一个表单.这意味着我希望将form(Textinput, Numberinput, Options...)中的所有选项Formlayout与captionto left和inputthe右对齐.

这适用于其他组件但不适用CheckBox.它总是将其标题放在Box的右侧,这会破坏我的表单的视觉顺序.

import java.util.Scanner;

public class JavaApplication10 {

    public static void main(String[] args) {
        Scanner reader = new Scanner(System.in);
        while (true) {
            int read = Integer.parseInt(reader.nextLine());
            if (read == 0) {
                break;
            }
            int sum;
            sum = read;
            System.out.println("Sum now: " + sum);
            while (true) {
                int read1 = Integer.parseInt(reader.nextLine());
                if (read1 != 0) {
                    sum += read1;

                    System.out.println("Sum now: " + sum);
                } else if (read1 == 0) {
                    System.out.println("Sum in the end: " + sum);
                    break;
                }
            }
        }
    } …

我想对字母数字值进行排序，就好像我的输入是

[a1, b1, c1, aa1, bb1, cc1, d1, e1]

输出应按此顺序- a1,b1,c1,d1,e1,aa1,bb1,cc1

我用过这个代码

public class coll {

    public static int comparator(String s1, String s2) {

        String[] pt1 = s1.split("((?<=[a-z])(?=[0-9]))|((?<=[0-9])(?=[a-z]))"); 
        String[] pt2 = s2.split("((?<=[a-z])(?=[0-9]))|((?<=[0-9])(?=[a-z]))"); 
//pt1 and pt2 arrays will have the string split in alphabets and numbers

        int i=0;
        if(Arrays.equals(pt1, pt2))
            return 0;
        else{
            for(i=0;i<Math.min(pt1.length, pt2.length);i++)
                if(!pt1[i].equals(pt2[i])) {
                    if(!isNumber(pt1[i],pt2[i])) {
                        if(pt1[i].compareTo(pt2[i])>0)
                            return 1;
                        else
                            return -1;
                    }
                    else {
                        int nu1 = Integer.parseInt(pt1[i]);
                        int nu2 = Integer.parseInt(pt2[i]);
                        if(nu1>nu2)
                            return 1;
                        else
                            return …