Home > Backend Development > Golang > How to solve 'cannot unmarshal...' error in golang?

How to solve 'cannot unmarshal...' error in golang?

WBOY
Release: 2023-06-24 20:01:59
Original
4332 people have browsed it

During the development process of Golang, we often use JSON for data exchange and serialization. However, when we try to deserialize JSON data into a structure, we sometimes encounter "cannot unmarshal..." errors. This error message is cryptic, difficult to understand, and confusing. This article will help you understand the cause of this error and provide several solutions.

Solution 1: Check the type of the structure member variables

The "cannot unmarshal..." error is usually caused by a mismatch between the value in the JSON string and the type of the structure member variable. . To avoid this error, we should ensure that when declaring the structure, the type of each member variable matches the corresponding value in JSON.

For example, we define the following structure:

type Person struct {
    Name string
    Age  int
}
Copy after login

If we try to deserialize the following JSON into a structure, we will encounter the "cannot unmarshal..." error:

{
    "Name": "Lucy",
    "Age": "28"
}
Copy after login

This is because the Age value type in JSON is a string and cannot match the Age member variable type int in the structure. We need to define the type of Age as string to deserialize JSON data correctly.

Solution 2: Check the tag of the structure member variable

Normally, if we define a tag (tag) for the field in the structure member variable, then the tag information will be reflected in the Guides JSON data during serialization, identifying the name and other details of each field. Therefore, when deserializing JSON, if the tag information does not match the attribute name in the JSON data, this error may also occur.

For example, we define the following structure:

type Person struct {
    Name string `json:"PersonName"`
    Age  int    `json:"PersonAge"`
}
Copy after login

If we try to deserialize the following JSON into a structure, we will also encounter the "cannot unmarshal..." error:

{
    "Name": "Lucy",
    "Age": 28
}
Copy after login

This is because we defined labels for the Name and Age fields in the structure, and the field names defined in the labels are inconsistent with the attribute names in the JSON data. Therefore, we need to change the attribute name in the JSON data to the tag name defined in the structure to deserialize it correctly.

Solution 3: Use golang’s own json package for annotation

Another method is to use golang’s own json package to prevent “cannot unmarshal…” errors through annotations . This method requires us to add json tags to each field when defining the structure, as shown below:

type Person struct {
    Name string `json:"name"`
    Age int `json:"age"`
}
Copy after login

When deserializing JSON, we need to use the json.Unmarshal method in the code and add JSON data as input. In this way, JSON data can be deserialized correctly, for example:

p := &Person{}
jsonData := []byte(`{"name":"John","age":18}`)
err := json.Unmarshal(jsonData, p)
if err != nil {
    fmt.Println("Unmarshal error: ", err)
}
fmt.Printf("Name: %s, Age: %d", p.Name, p.Age)
Copy after login

Summary

In most cases, the "cannot unmarshal..." error is usually due to the types and structures in the JSON data This is caused by the type mismatch of the body member variables or the mismatch between the tag information and the attribute names in the JSON data. By carefully checking these issues, we can avoid this error. If you want to better solve this error, please use Golang's own json package for annotation management.

The above is the detailed content of How to solve 'cannot unmarshal...' error in golang?. For more information, please follow other related articles on the PHP Chinese website!

Related labels:
source:php.cn
Statement of this Website
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn
Popular Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template