How to build a JSON string with Bash variables

Constructing JSON strings in Bash is a common task, especially when interacting with APIs via scripts. Here, I'll explain with an example how to construct a JSON string using Bash variables.

Suppose we need to create a JSON object representing a user, containing the user's name and age. First, we define two variables to store the user's name and age:

bash
name="John Doe"
age=30

Next, we need to construct a JSON string. In Bash, we can use double quotes to enclose strings and the ${variable} syntax to insert variables. A simple way to construct JSON strings is to directly use string concatenation:

bash
json_string="{\"name\": \"${name}\", \"age\": ${age}}"

Here, we use escape characters \ to ensure that quotes are correctly interpreted as part of the string. Thus, the json_string variable contains the following content:

json
{"name": "John Doe", "age": 30}

However, in practical applications, direct string concatenation for JSON construction can lead to issues, such as when variable content contains special characters. To avoid these issues, we can use a command-line JSON processor like jq to safely generate JSON strings:

bash
json_string=$(jq -n \
                  --arg name "$name" \
                  --argjson age $age \
                  '{name: $name, age: $age}')

Here, jq -n indicates creating a new JSON object, --arg is used to create a JSON string, and --argjson is used to create a JSON number. The final json_string variable will contain the same content:

json
{"name": "John Doe", "age": 30}

The benefit of using jq is that it can handle special characters and ensures the generated JSON is correctly formatted. This is particularly useful for scenarios where scripts need to handle complex data structures.