用里面的变量在 Tcl 中打开文件

Question

I want to open a file called filelist.txt which contains only the string ${PATHFILE}/test.txt , read the line and open the file test.txt . The file test.txt is present inside the folder ~/testfile .
Consider this sample code:

#!/usr/bin/env tclsh

set PATHFILE "~/testfile"

set fp [open "filelist.txt" r]
set lines [split [read $fp] "\n"]
close $fp   

foreach line $lines {
    set fp1 [open $line r]
    close $fp1
}

The problem is that it seems that the "open" command cannot find the PATHFILE variable and i get this error:

couldn't open "${PATHFILE}/test.txt": no such file or directory

If i try to open the file with set fp1 [open "${PATHFILE}/test.txt" r] i don't have any errors.

Answer 1

Yes, you can use the TCL subst command to evaluate the PATHFILE variable. Note that you might still have an issue with the tilde ~ - it may be better to use full path names.

#!/usr/bin/env tclsh

set PATHFILE "~/testfile"

set fp [open "filelist.txt" r]
set lines [split [read $fp] "\n"]
close $fp   

foreach line $lines {
    set fp1 [open [subst $line] r]
    close $fp1
}