Per-row labels with #[umbral(str = "...")]

Django models have __str__ — the one method that turns a row into a human string ("Ada Lovelace", not User #7). umbral's equivalent is a struct-level template:

#[derive(Debug, Clone, sqlx::FromRow, serde::Serialize, serde::Deserialize, Model)]
#[umbral(table = "person", str = "{first_name} {last_name}")]
pub struct Person {
    pub id: i64,
    pub first_name: String,
    pub last_name: String,
}

Each {field} placeholder is substituted with that row's value, so a Person row renders as "Ada Lovelace". The admin uses it wherever it labels a single object — foreign-key and many-to-many chips, related-object pickers, and detail titles — so a ForeignKey<Person> shows the name instead of a bare id.

The rules

• Any field name works: str = "{title} ({year})", str = "#{id} — {name}".
• Numbers and booleans stringify; a null or absent field becomes empty.
• {{ and }} are literal braces.

Fallback when there's no template

Without #[umbral(str = ...)], the admin falls back — in order — to the field marked #[umbral(string)], then the first text column, then the primary key. So a single-field label is still just #[umbral(string)] on that field; reach for str when a label needs more than one field concatenated.

Using it yourself

ModelMeta::render_str(&row) renders the template against a JSON row (returning None when the model declares no template), so any code holding a ModelMeta and a row can produce the same label the admin does.