Spring Expression直接访问Selection中的映射键

Question

考虑我已经将JSON字符串解析为Map<String, Object>，这是任意嵌套的

我的JSON看起来像：

{
  "root": [
   {"k":"v1"},
   {"k":"v2"}
  ]
}

我尝试了表达式root.?[k == 'v1']，但收到了以下错误：

EL1008E: Property or field 'k' cannot be found on object of type 'java.util.LinkedHashMap' - maybe not public?

Answer 1

评估上下文需要MapAccessor：

public static void main(String[] args) throws Exception {
    String json = "{\n" +
            "  \"root\": [\n" +
            "   {\"k\":\"v1\"},\n" +
            "   {\"k\":\"v2\"}\n" +
            "  ]\n" +
            "}";
    ObjectMapper mapper = new ObjectMapper();
    Object root = mapper.readValue(json, Object.class);
    Expression expression = new SpelExpressionParser().parseExpression("root.?[k == 'v1']");
    StandardEvaluationContext ctx = new StandardEvaluationContext();
    ctx.addPropertyAccessor(new MapAccessor());
    System.out.println(expression.getValue(ctx, root));
}

结果：

[{k=v1}]

如果没有MapAccessor，则需要

"['root'].?[['k'] == 'v1']"

MapAccessor仅适用于不包含句点的地图键。