Malformed mapping no type name found – How to solve this Elasticsearch exception

Opster Team

August-23, Version: 6.8-8.9

Before you dig into reading this guide, have you tried asking OpsGPT what this log means? You’ll receive a customized analysis of your log.

Try OpsGPT now for step-by-step guidance and tailored insights into your Elasticsearch operation.

Briefly, this error occurs when Elasticsearch encounters an incorrect or incomplete mapping structure. Mapping defines the data types for fields, and if a type name is missing, Elasticsearch cannot process the data correctly. To resolve this issue, you can review and correct your mapping structure, ensuring that each field has a defined type. Alternatively, you may need to update your Elasticsearch version if it no longer supports types in mapping, as newer versions only allow single-type indices.

For a complete solution to your to your search operation, try for free AutoOps for Elasticsearch & OpenSearch . With AutoOps and Opster’s proactive support, you don’t have to worry about your search operation – we take charge of it. Get improved performance & stability with less hardware.

This guide will help you check for common problems that cause the log ” malformed mapping; no type name found ” to appear. To understand the issues related to this log, read the explanation below about the following Elasticsearch concepts: index.

Log Context

Log “malformed mapping; no type name found” class name is We extracted the following from Elasticsearch source code for those seeking an in-depth context :

 Mapping parse(@Nullable String type; CompressedXContent source) throws MapperParsingException {
 Objects.requireNonNull(source; "source cannot be null");
 Map mapping = XContentHelper.convertToMap(source.compressedReference(); true; XContentType.JSON).v2();
 if (mapping.isEmpty()) {
 if (type == null) {
 throw new MapperParsingException("malformed mapping; no type name found");
 } else {
 String rootName = mapping.keySet().iterator().next();
 if (type == null || type.equals(rootName) || documentTypeResolver.apply(type).equals(rootName)) {
 type = rootName;


How helpful was this guide?

We are sorry that this post was not useful for you!

Let us improve this post!

Tell us how we can improve this post?

Get expert answers on Elasticsearch/OpenSearch