在開發過程中,在對list集合的排序中遇到了點小阻礙。記錄之,與君共勉。
我們先來看看,Collections.sort()的簡單用法,代碼:
package com.wh.util; import org.slf4j.Logger; import org.slf4j.LoggerFactory; import java.util.ArrayList; import java.util.Collections; import java.util.List; public class Test { private static final Logger LOGGER = LoggerFactory.getLogger(Test.class); public static void main(String[] args) { List<String> lists = new ArrayList<String>(); lists.add("1"); lists.add("5"); lists.add("3"); LOGGER.info("lists 排序前:" + lists.toString()); //排序,直接調用sort方法排序,排序方式是自燃排序,即升序排序 Collections.sort(lists); LOGGER.info("lists 排序后:" + lists.toString()); } }
運行結果:
完美排序。但是在實際開發中排序的集合遠比這個復雜,我們來看簡單的模擬。
package com.wh.bean; import java.io.Serializable; /** * Created by WH on 2016/9/17. */ public class TestA implements Serializable{ private static final long serialVersionUID = -8968350748602548887L; private String name; //變量類型必須要為int的封裝類型,因為只有是封裝類型才可以調用Integer的方法 private Integer order; public String getName() { return name; } public void setName(String name) { this.name = name; } public Integer getOrder() { return order; } public void setOrder(int order) { this.order = order; } @Override public String toString() { return "TestA{" + "name='" + name + '\'' + ", order=" + order + '}'; } }
在調用Collections.sort()給listA排序中,編譯不通過,報錯了。這就是當時我開發中遇到的小阻礙,為什么編譯不通過呢,前面的演示中為什么可以排序呢?這是因為lists中的String本身含有compareTo方法,所以可以直接調用sort方法,所以要讓類TestA實現Comparable方法,當時參考了網上資料(這里)。解決代碼如下:
package com.wh.util; import com.wh.bean.TestA; import org.slf4j.Logger; import org.slf4j.LoggerFactory; import java.util.ArrayList; import java.util.Collections; import java.util.Comparator; import java.util.List; public class Test { private static final Logger LOGGER = LoggerFactory.getLogger(Test.class); public static void main(String[] args) { List<TestA> listA = new ArrayList<TestA>(); //TestA 即是上面類 TestA a1 = new TestA(); a1.setName("張大胖"); a1.setOrder(1); listA.add(a1); TestA a5 = new TestA(); a5.setName("王曉梅"); a5.setOrder(5); listA.add(a5); TestA a3 = new TestA(); a3.setName("劉曉彤"); a3.setOrder(3); listA.add(a3); LOGGER.info("listA 排序前:" + listA.toString()); Collections.sort(listA, new Comparator<TestA>() { @Override public int compare(TestA o1, TestA o2) { //升序 return o1.getOrder().compareTo(o2.getOrder()); } }); LOGGER.info("listB 升序排序后:" +listA.toString()); Collections.sort(listA, new Comparator<TestA>() { @Override public int compare(TestA o1, TestA o2) { //降序 return o2.getOrder().compareTo(o1.getOrder()); } }); LOGGER.info("listB 降序排序后:" +listA.toString()); } }
運行結果:
這里可以實現降序和升序排序。除了這個解決方法還有另外一個解決方法:
package com.wh.bean; import java.io.Serializable; /** * Created by WH on 2016/9/17. */ //實現Comparable public class TestB implements Serializable, Comparable<TestB> { private static final long serialVersionUID = -2872189514594810657L; private String name; private Integer order; public String getName() { return name; } public void setName(String name) { this.name = name; } public int getOrder() { return order; } public void setOrder(Integer order) { this.order = order; } @Override public int compareTo(TestB o) {
//升序 return this.order.compareTo(o.getOrder()); } @Override public String toString() { return "TestB{" + "name='" + name + '\'' + ", order=" + order + '}'; } }
package com.wh.bean; import java.io.Serializable; /** * Created by WH on 2016/9/17. */ public class TestB implements Serializable, Comparable<TestB> { private static final long serialVersionUID = -2872189514594810657L; private String name; private Integer order; public String getName() { return name; } public void setName(String name) { this.name = name; } public int getOrder() { return order; } public void setOrder(Integer order) { this.order = order; } @Override public int compareTo(TestB o) { return this.order.compareTo(o.getOrder()); } @Override public String toString() { return "TestB{" + "name='" + name + '\'' + ", order=" + order + '}'; } }
運行結果:
就這樣解決了問題。