问题
使用HuTool的JSONUtil工具类时遇到JSONUtil.parse()无法解析字符串的问题
@Test
public void test1() {
String str = "1234";
System.out.println(JSONUtil.parse(str).toStringPretty());
}
图1 parse方法
parse方法传入一个Object类型对象, 支持String, Array数组, Collection集合, 和Bean对象, 但在解析字符串时抛出异常cn.hutool.json.JSONException: A JSONObject text must begin with '{' at 1 [character 2 line 1]
cn.hutool.json.JSONException: A JSONObject text must begin with '{' at 1 [character 2 line 1]
at cn.hutool.json.JSONTokener.syntaxError(JSONTokener.java:396)
at cn.hutool.json.JSONObject.init(JSONObject.java:736)
at cn.hutool.json.JSONObject.init(JSONObject.java:723)
at cn.hutool.json.JSONObject.init(JSONObject.java:683)
at cn.hutool.json.JSONObject.<init>(JSONObject.java:202)
at cn.hutool.json.JSONObject.<init>(JSONObject.java:178)
at cn.hutool.json.JSONObject.<init>(JSONObject.java:160)
at cn.hutool.json.JSONObject.<init>(JSONObject.java:143)
at cn.hutool.json.JSONUtil.parseObj(JSONUtil.java:88)
at cn.hutool.json.JSONUtil.parse(JSONUtil.java:221)
at cn.hutool.json.JSONUtil.parse(JSONUtil.java:196)
at org.casey.basic.hutool.JsonTest.test1(JsonTest.java:22)
at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62)
at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
at java.lang.reflect.Method.invoke(Method.java:498)
at org.junit.runners.model.FrameworkMethod$1.runReflectiveCall(FrameworkMethod.java:59)
at org.junit.internal.runners.model.ReflectiveCallable.run(ReflectiveCallable.java:12)
at org.junit.runners.model.FrameworkMethod.invokeExplosively(FrameworkMethod.java:56)
at org.junit.internal.runners.statements.InvokeMethod.evaluate(InvokeMethod.java:17)
at org.junit.runners.ParentRunner$3.evaluate(ParentRunner.java:306)
at org.junit.runners.BlockJUnit4ClassRunner$1.evaluate(BlockJUnit4ClassRunner.java:100)
at org.junit.runners.ParentRunner.runLeaf(ParentRunner.java:366)
at org.junit.runners.BlockJUnit4ClassRunner.runChild(BlockJUnit4ClassRunner.java:103)
at org.junit.runners.BlockJUnit4ClassRunner.runChild(BlockJUnit4ClassRunner.java:63)
at org.junit.runners.ParentRunner$4.run(ParentRunner.java:331)
at org.junit.runners.ParentRunner$1.schedule(ParentRunner.java:79)
at org.junit.runners.ParentRunner.runChildren(ParentRunner.java:329)
at org.junit.runners.ParentRunner.access$100(ParentRunner.java:66)
at org.junit.runners.ParentRunner$2.evaluate(ParentRunner.java:293)
at org.junit.runners.ParentRunner$3.evaluate(ParentRunner.java:306)
at org.junit.runners.ParentRunner.run(ParentRunner.java:413)
at org.junit.runner.JUnitCore.run(JUnitCore.java:137)
at com.intellij.junit4.JUnit4IdeaTestRunner.startRunnerWithArgs(JUnit4IdeaTestRunner.java:68)
at com.intellij.rt.junit.IdeaTestRunner$Repeater.startRunnerWithArgs(IdeaTestRunner.java:33)
at com.intellij.rt.junit.JUnitStarter.prepareStreamsAndStart(JUnitStarter.java:230)
at com.intellij.rt.junit.JUnitStarter.main(JUnitStarter.java:58)
原因
我的失误, 一时忘记了JSON是键值对的格式, 试想, 如果没有key光有value如何构成JSON对象的属性呢
解决
构造一个字典来进行解析, 构造键值对
@Test
public void test2() {
String str = "1234";
Dict dict = Dict.create();
dict.put("data", str);
System.out.println(JSONUtil.parse(dict).toStringPretty());
}
{
"data": "1234"
}
Process finished with exit code 0