A bash snippet to get the directory a script is located in

See https://stackoverflow.com/a/246128/3561275 for details about how
this works.

Author: omajid

snippets/sh-mode/script-dir

@@ -0,0 +1,14 @@
+# -*- mode: snippet -*-
+# name: the currently executing/sourced script's directory
+# key: script-dir
+# --
+# See https://stackoverflow.com/a/246128/3561275
+SOURCE="\${BASH_SOURCE[0]}"
+while [ -h "\$SOURCE" ]; do # resolve \$SOURCE until the file is no longer a symlink
+  DIR="\$( cd -P "\$( dirname "\$SOURCE" )" >/dev/null 2>&1 && pwd )"
+  SOURCE="\$(readlink "\$SOURCE")"
+  [[ \$SOURCE != /* ]] && SOURCE="\$DIR/\$SOURCE" # if \$SOURCE was a relative symlink, we need to resolve it relative to the path where the symlink file was located
+done
+DIR="\$( cd -P "\$( dirname "\$SOURCE" )" >/dev/null 2>&1 && pwd )"
+
+$0