get_tables_from_sql.lazy_base_remote_query not working

[willshen99]

I installed RPresto from current github repo. My dbplyr version is 2.5.0.

tbl(con, 'table_name') is not working. Full backtrace:

Error in `as_table_path()`:
! `con` is absent but must be supplied.
---
Backtrace:
     ▆
  1. ├─base (local) `<fn>`(x)
  2. └─dbplyr:::print.tbl_sql(x)
  3.   ├─dbplyr:::cat_line(format(x, ..., n = n, width = width, n_extra = n_extra))
  4.   │ ├─base::cat(paste0(..., "\n"), sep = "")
  5.   │ └─base::paste0(..., "\n")
  6.   ├─base::format(x, ..., n = n, width = width, n_extra = n_extra)
  7.   └─pillar:::format.tbl(x, ..., n = n, width = width, n_extra = n_extra)
  8.     └─pillar:::format_tbl(...)
  9.       └─pillar::tbl_format_setup(...)
 10.         ├─pillar:::tbl_format_setup_dispatch(...)
 11.         └─pillar:::tbl_format_setup.tbl(...)
 12.           └─pillar:::df_head(x, n + 1)
 13.             ├─base::as.data.frame(head(x, n))
 14.             └─dbplyr:::as.data.frame.tbl_sql(head(x, n))
 15.               ├─base::as.data.frame(collect(x, n = n))
 16.               ├─dplyr::collect(x, n = n)
 17.               └─RPresto:::collect.tbl_presto(x, n = n)
 18.                 └─dbplyr::db_sql_render(dbplyr::remote_con(x), x)
 19.                   ├─dbplyr::db_sql_render(con, sql, ..., sql_options = sql_options)
 20.                   └─RPresto:::db_sql_render.PrestoConnection(con, sql, ..., sql_options = sql_options)
 21.                     └─RPresto:::find_cte_tables_from_lazy_query(con, sql$lazy_query)
 22.                       ├─base::unique(get_tables_from_sql(sql))
 23.                       ├─RPresto:::get_tables_from_sql(sql)
 24.                       └─RPresto:::get_tables_from_sql.lazy_select_query(sql)
 25.                         ├─RPresto:::get_tables_from_sql(query$x)
 26.                         └─RPresto:::get_tables_from_sql.lazy_base_remote_query(query$x)
 27.                           └─(utils::getFromNamespace("table_path_name", "dbplyr"))(query$x)
 28.                             └─dbplyr::as_table_path(x, con)

[willshen99]

The issue is due to this line

  if (inherits(query$x, "dbplyr_table_path")) { # dbplyr >= 2.5.0
    utils::getFromNamespace("table_path_name", "dbplyr")(query$x)

as_table_path should take both x and con.