Question

我试图将String转换为Json Array（无名称/标签）并遍历结果数组。.但是，当我调试代码时，结果json数组显示为null。我找不到导致此问题的原因。

下面是我要解析为Json数组的给定字符串。

String jsonString="[ {"
            + "\"TimeStamp\": \"20:10\","
            + "\"PipeTemp\": \"31.5\","
            + "\"ChocolateMixingTemp\": \"25\","
            + "\"WaterCoolingTemp\": \"5\","
            + "\"WaterMixingTemp\": \"0\","
            + "\"WaterHeatingTemp\": \"0\",},
            +{"
            + "\"TimeStamp\": \"20:00\","
            + "\"PipeTemp\": \"35.5\","
            + "\"ChocolateMixingTemp\": \"28\","
            + "\"WaterCoolingTemp\": \"3\","
            + "\"WaterMixingTemp\": \"15\","
            + "\"WaterHeatingTemp\": \"0\","
            +}]";

我将上面的字符串传递给Java Array实例，然后使用for循环对其进行迭代以生成PDF表。但是当我尝试检查其大小时，我看到Json Array显示为null。

JSONArray jsonArray = new JSONArray(jsonString);

 JSONObject jsonObject;
 for(int n = 0; n < jsonArray.length(); n++)
{
 table.addCell(createCell( (String) jsonObject.get("TimeStamp"),Element.ALIGN_CENTER)); 
             table.addCell(createCell( (String) jsonObject.get("PipeTemp"),Element.ALIGN_CENTER)); 
             table.addCell(createCell( (String) jsonObject.get("ChocolateMixingTemp"), Element.ALIGN_CENTER)); 
             table.addCell(createCell( (String) jsonObject.get("WaterCoolingTemp"),Element.ALIGN_CENTER)); 
             table.addCell(createCell( (String) jsonObject.get("WaterMixingTemp"),Element.ALIGN_CENTER)); 
jsonObject.get("WaterHeatingTemp"),Element.ALIGN_CENTER)); 

}

有人可以帮助我确定为什么数组显示为空吗？我还尝试使用仅包含两个键值对的简单字符串进行测试。但是得到了相同的结果。

Answer 1

您的代码错误。尝试以下代码来测试JSON数组：

String jsonString = "[ {"
            + "\"TimeStamp\": \"20:10\","
            + "\"PipeTemp\": \"31.5\","
            + "\"ChocolateMixingTemp\": \"25\","
            + "\"WaterCoolingTemp\": \"5\","
            + "\"WaterMixingTemp\": \"0\","
            + "\"WaterHeatingTemp\": \"0\",},"
            + "{"
            + "\"TimeStamp\": \"20:00\","
            + "\"PipeTemp\": \"35.5\","
            + "\"ChocolateMixingTemp\": \"28\","
            + "\"WaterCoolingTemp\": \"3\","
            + "\"WaterMixingTemp\": \"15\","
            + "\"WaterHeatingTemp\": \"0\","
            + "}]";

JSONArray jsonArray = new JSONArray(jsonString);

System.out.println(jsonArray);

for (int n = 0; n < jsonArray.length(); n++) {
    JSONObject jsonObject = jsonArray.getJSONObject(n);
    System.out.println((String) jsonObject.get("TimeStamp"));
}

Answer 2

首先，您的json字符串格式不正确，请参见以下格式：

String s="[ {"
                + "\"TimeStamp\": \"20:10\","
                + "\"PipeTemp\": \"31.5\","
                + "\"ChocolateMixingTemp\": \"25\","
                + "\"WaterCoolingTemp\": \"5\","
                + "\"WaterMixingTemp\": \"0\","
                + "\"WaterHeatingTemp\": \"0\",},"
                + "{"
                + "\"TimeStamp\": \"20:00\","
                + "\"PipeTemp\": \"35.5\","
                + "\"ChocolateMixingTemp\": \"28\","
                + "\"WaterCoolingTemp\": \"3\","
                + "\"WaterMixingTemp\": \"15\","
                + "\"WaterHeatingTemp\": \"0\","
                + "}]";

第二，我看不到您的jsonObject被初始化。如果要使用json中的值，请使用jsonArray.get（index）方法。

在Java中从String转换为Json Array时获得空Json Array

2 个答案: