Only one type can be output when there are multiple structs with the same name.

[lzyor]

If more than one struct with the same name exists at the same time, only the type of one of them can be output. When structs are defined in different files, sometimes one will be randomly selected for output.

Adding #[serde(rename = “DataA”)] still does this.

use serde::{Deserialize, Serialize};

#[typeshare::typeshare]
#[derive(Serialize, Deserialize)]
#[serde(rename = "DataA")]
pub struct Data {
    pub t: String,
    pub a: u8,
}

pub mod another {
    use serde::{Deserialize, Serialize};

    #[typeshare::typeshare]
    #[derive(Serialize, Deserialize)]
    #[serde(rename = "DataB")]
    pub struct Data {
        pub t: String,
        pub s: String,
    }
}

output:

/*
 Generated by typeshare 1.13.2
*/

export interface DataA {
	t: string;
	a: number;
}

// There should be `DataB` ?

Some libraries, such as sea_orm require model definitions to be named Model. this results in only one of the definitions being output.

[darrell-roberts]

If your ok with scoping types with the same name into separate crates then you could use the -d, --output-folder option which generates multiple files for each crate. This feature was added in part to support name spacing beyond a single name space.