Spring Expression 直接访问 Selection 中的映射键

Question

考虑到我已经将一个 JSON 字符串解析为一个 Map<String, Object>，它是任意嵌套的

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}]

没有MapAccessor，你需要

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

A MapAccessor 仅适用于不包含句点的地图键。

Spring Expression 直接访问 Selection 中的映射键

Spring Expression Directly access map keys in Selection

spring

spring-el