我想像这样转换String
myString:
[ ["cd",5,6,7], ["rtt",55,33,12], ["by65",87,87,12] ]
变成一个 ArrayList<CustomClass>
哪里CustomClass
有构造函数:
public CustomClass (String name, int num1, int num2, int num3)
我第一次尝试创建ArrayList
的Strings
:
List<String> List = new ArrayList<String>(Arrays.asList(myString.split("[")));
不是为我工作...
我如何得到这样的东西:
List - {CustomClass,CustomClass,CustomClass,CustomClass}
第一的 CustomClass = CustomClass.name="cd" , CustomClass.num1=5,CustomClass.num2=7...
第二 CustomClass = CustomClass.name="rtt",CustomClass.num1=55,CustomClass.num2=55...
等等...
你可以做类似的事情。如果您不能保证字符串格式,则可能必须添加其他检查以检查拼接数组的长度和索引。
import java.util.ArrayList;
import java.util.List;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
class CustomClass {
String name;
int num1;
int num2;
int num3;
public CustomClass(String name, int num1, int num2, int num3) {
super();
this.name = name;
this.num1 = num1;
this.num2 = num2;
this.num3 = num3;
}
}
public class Sample {
public static void main(String[] args) {
String str = "[ [\"cd\",5,6,7], [\"rtt\",55,33,12], [\"by65\",87,87,12] ]";
Pattern p = Pattern.compile("\\[(.*?)\\]");
Matcher m = p.matcher(str.substring(1));
List<CustomClass> customList = new ArrayList<CustomClass>();
while (m.find()) {
String[] arguments = m.group(1).split(",");
customList.add(new CustomClass(arguments[0],
Integer.parseInt(arguments[1]),
Integer.parseInt(arguments[2]),
Integer.parseInt(arguments[3])));
}
}
}
Gson解决方案
public static void main(String[] args) {
String json = "[ [\"cd\",5,6,7], [\"rtt\",55,33,12], [\"by65\",87,87,12] ]";
List<CustomClass> customList = new ArrayList<CustomClass>();
String[][] data = new Gson().fromJson(json, String[][].class);
for (String[] strArray : data){
customList.add(new CustomClass(strArray[0],
Integer.parseInt(strArray[1]),
Integer.parseInt(strArray[2]),
Integer.parseInt(strArray[3])));
}
System.out.println(customList);
}
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