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Table of Contents
How to parse XML data
How to generate XML data
Tips for dealing with complex or irregular structures
Frequently Asked Questions and Notes
Home Backend Development Golang How to work with xml in golang

How to work with xml in golang

Jul 09, 2025 am 01:08 AM

Parsing and generating XML data can be implemented through structure mapping in Go language. 1. To parse XML, you need to define the corresponding structure and convert it with xml.Unmarshal(). The field name matches the label and the first letter is capitalized; 2. When generating XML, you can beautify the output with xml.Marshal() or xml.MarshalIndent(). The default root element is the structure name; 3. When dealing with complex structures, you can use pointer types to deal with missing fields, or use a general structure to process different child nodes; 4. Pay attention to common problems such as namespace, performance optimization, and case sensitivity.

How to work with xml in golang

Processing XML data is actually quite straightforward in Go language. As long as you master the basic structure and several key functions, it is still very convenient to use. encoding/xml package in the Go standard library has provided many practical functions, both parsing and generating XML.

How to work with xml in golang

Here are a few common scenarios from the perspective of actual use to help you get started quickly.

How to work with xml in golang

How to parse XML data

The most common way to parse XML is to first define a structure, and the field name corresponds to the tag name of the XML. Then use xml.Unmarshal() to convert the original data into a structure.

Let's give a simple example:

How to work with xml in golang
 type Person struct {
    Name string `xml:"name"`
    Age int `xml:"age"`
}

If you have an XML string:

 <person><name>Alice</name><age>30</age></person>

This is how to parse:

 var p Person
err := xml.Unmarshal(data, &p)

A few points to note:

  • Structure fields to be exported (first letter capitalization)
  • The tag should be written accurately, such as xml:"name" corresponding to <name> tag
  • If there is a nested structure in XML, the structure must also be nested accordingly

If the XML structure is complex or uncertain, you can consider using map[string]interface{} or manually traversing the token to parse.


How to generate XML data

Conversely, if you want to convert a structure into an XML string, you can use xml.Marshal() or xml.MarshalIndent() to beautify the output format.

For example:

 p := Person{Name: "Bob", Age: 25}
data, _ := xml.Marshal(p)

The default output is not indented in a new line. If you want it to be easier to read, you can use:

 data, _ := xml.MarshalIndent(p, "", " ")

The result is probably like this:

 <Person>
  <Name>Bob</Name>
  <Age>25</Age>
</Person>

Notice:

  • The output root element name is the name of the structure type by default.
  • You can customize the tag name by setting xml:"mytag" tag
  • If you want to add an XML declaration (for example, <?xml version="1.0" encoding="UTF-8"?> ), you need to manually splice it yourself

Tips for dealing with complex or irregular structures

Sometimes the XML structure is not fixed, such as some fields may not exist, or the tag sequence is confused. This is where some additional processing is needed.

For example, if a field may be missing, it can be declared as a pointer type in the structure:

 type Item struct {
    ID int `xml:"id"`
    Desc *string `xml:"description"`
}

If there is no <description> in XML, then this field will be nil and there will be no error.

Another situation is that there are multiple child nodes under the same tag, but their names are different, such as:

 <items><itemA>One</itemA><itemB>Two</itemB></items>

In this case, you can define a general structure, or use struct{} to skip the specific structure and extract only the required parts.


Frequently Asked Questions and Notes

  • Namespace : XML often has a namespace, such as <name> . Go's standard library supports handling of namespaces, but requires manual processing of prefixes and URIs.
  • Performance issues : For very large XML files, parsing to memory at one time may not be efficient enough. At this time, it is recommended to use a streaming parser, such as Decoder.Token() method to process one line by line.
  • Case sensitive : The structure fields of Go are case sensitive, so be sure to make sure that the tag and XML tags exactly match.
  • Comments and special nodes : The standard library does not support retaining comments, CDATA, etc. If you have this requirement, you may need third-party libraries or expand them yourself.

Basically that's it. Although XML is not as popular as JSON, it cannot be avoided in some system docking. By mastering the structure mapping and analysis methods, you can easily deal with most scenarios.

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