Cannot deserialize value of type `java.util.LinkedHashMap` from Object value (token `JsonToken.START_OBJECT`): what it means and how to fix it

TL;DR: Validate locally, fix the first real error, validate again (no upload).

Troubleshoot Cannot deserialize value of type `java.util.LinkedHashMap` from Object value (token `JsonToken.START_OBJECT`) quickly and validate JSON locally (no upload).

What the error means

Cannot deserialize value of type `java.util.LinkedHashMap` from Object value (token `JsonToken.START_OBJECT`) usually means the JSON was parsed, but the value at some path does not match the type/schema your code expects (for example: array vs object, number vs string, null vs non-nullable).

Most common real-world causes

  • Your code expects a different type (number vs string, bool vs string, object vs array).
  • A field is null/missing but your type is non-nullable (or your code assumes it is always present).
  • The API response shape changed (or differs by environment / feature flags).
  • You are deserializing the wrong endpoint or the response is an error payload with a different schema.
  • Date/UUID/IP formats don't match what the deserializer expects.

Fast debugging steps

  • Parse into a generic JSON type first (Value/JToken/JsonElement/JsonNode) and inspect the actual shape at the failing path.
  • Confirm whether the field is an array vs object and whether scalars are strings vs numbers.
  • Add defensive handling for null/missing fields (optional types, defaults, or custom converters).
  • Log the raw response (or a redacted sample) and validate it locally to rule out truncation or hidden prefixes.
  • Update DTO/types to match the real response shape (or use a more flexible model for unstable fields).

Code example (java)

// Parse into JsonNode first to inspect shape/type mismatches
ObjectMapper mapper = new ObjectMapper();
JsonNode node = mapper.readTree(body);

// Example checks
JsonNode value = node.get("value");
if (value != null) {
  System.out.println(value.getNodeType());
  System.out.println(value.toString());
}

// Then deserialize into your DTO once the schema matches
// MyDto dto = mapper.treeToValue(node, MyDto.class);

Fix without uploading data

If the JSON contains sensitive data, validate and fix it locally. No Upload Tools runs 100% in your browser.

Workflow: validate -> fix the first error -> validate again -> export/convert.

FAQ

Does this mean the JSON is invalid? Not necessarily. Many of these errors happen when the JSON is valid, but your code expects a different type/shape.

What is the fastest fix? Parse into a generic JSON type first (Value/JToken/JsonElement/JsonNode), inspect the failing path, then update your DTO/type (or add a converter) to match the real response.

Privacy & Security
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Запрос из поиска

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  • Проверьте структуру и типы входных данных.
  • Найдите позицию ошибки и изолируйте минимальный пример.
  • Сверьте экранирование, разделители и кодировку.
  • Примените фикс и повторите проверку на реальном payload.

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Quick fix checklist

  • Reproduce the error on a minimal input.
  • Check type/format and field mapping.
  • Apply the smallest safe fix.
  • Validate on production-like payload.