php array 转json及java 转换 json数据格式操作示例
人气:0本文实例讲述了php array 转json及java 转换 json数据格式操作。分享给大家供大家参考,具体如下:
php array 转json 数据
$arr = array( "demo1" => "my demo1", "demo2" => array( "demo2_demo1"=>"aaaaaaaa", "demo2_demo2"=>"bbbbbbbb", "demo2_demo3"=>array( "demo2_demo3_demo1"=>"ccccccc" ) ), "demo3" => 22 ); $json_str = json_encode($arr,true); $arr = json_decode($json_str,true); var_dump($json_str); var_dump($arr); //查看结果
运行结果:
string(137) "{"demo1":"my demo1","demo2":{"demo2_demo1":"aaaaaaaa","demo2_demo2":"bbbbbbbb","demo2_demo3":{"demo2_demo3_demo1":"ccccccc"}},"demo3":22}"
array(3) {
["demo1"]=>
string(8) "my demo1"
["demo2"]=>
array(3) {
["demo2_demo1"]=>
string(8) "aaaaaaaa"
["demo2_demo2"]=>
string(8) "bbbbbbbb"
["demo2_demo3"]=>
array(1) {
["demo2_demo3_demo1"]=>
string(7) "ccccccc"
}
}
["demo3"]=>
int(22)
}
java json数据格式转换依赖包
commons-beanutils-1.8.3.jar,
commons-collections-3.2.1.jar,
commons-lang-2.6.jar,
commons-logging-1.1.1.jar,
ezmorph-1.0.6.jar,
json-lib-2.4-jdk15.jar,
添加至构建路径。
package Main; import net.sf.json.JSONObject; import net.sf.json.JSONArray; import java.util.Map; import java.util.HashMap; import java.util.List; import java.util.ArrayList; import Main.Demo1; public class Index { public static void main(String[] args) { index4(); } /** * 简单的simple * */ public static void index1() { JSONObject json = new JSONObject(); json.element("name","谭勇"); json.element("age",22); System.out.println(json.toString()); } /** * Map 数据转json * */ public static void index2() { JSONObject json = new JSONObject(); Map<String,String> map = new HashMap<String,String>(); map.put("name", "谭勇"); map.put("age", "22"); json.accumulateAll(map); System.out.println(json.toString()); } /** * List<Map> 转json * */ public static void index3() { JSONArray arr = new JSONArray(); List<Map<String,String>> list = new ArrayList<Map<String,String>>(); list.add(getMap("name","谭勇")); list.add(getMap("age","22")); arr.addAll(list); System.out.println(arr.toString()); } /** * 对象转Json * */ public static void index4() { Demo1 demo1 = new Demo1(); demo1.setName("谭勇"); demo1.setAge(22); JSONObject json = new JSONObject(); JSONArray arr = new JSONArray(); json.element("demo1",demo1); arr.add(demo1); //json.containsKey(demo1); System.out.println(json.toString()); System.out.println(arr.toString()); } private static Map<String,String> getMap(String key,String val) { Map<String,String> map = new HashMap<String,String>(); map.put(key, val); return map; } }
Demo1.java
package Main; public class Demo1 { private String name; private int age; public void setName(String name) { this.name = name; } public String getName() { return this.name; } public void setAge(int age) { this.age = age; } public int getAge() { return this.age; } }
希望本文所述对大家PHP程序设计有所帮助。
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