LibreOffice 25.2 Bantuan
Mengembalikan untai teks dari teks DBCS. Parameter menentukan posisi awal dan nomor karakter.
MIDB("Teks"; Mulai; Nomor_bita)
Teks adalah teks yang berisi karakter yang diekstrak.
Mulai adalah posisi pertama karakter dalam teks yang diekstrak.
Nomor_bita menentukan jumlah karakter yang akan dikembalikan dari teks, dalam bita.
=MIDB("ä¸ĺ›˝";1;0) returns "" (0 bytes is always an empty string).
=MIDB("ä¸ĺ›˝";1;1) returns " " (1 byte is only half a DBCS character and therefore the result is a space character).
=MIDB("ä¸ĺ›˝";1;2) returns "ä¸" (2 bytes constitute one complete DBCS character).
=MIDB("ä¸ĺ›˝";1;3) returns "ä¸ " (3 bytes constitute one and a half DBCS character; the last byte results in a space character).
=MIDB("ä¸ĺ›˝";1;4) returns "ä¸ĺ›˝" (4 bytes constitute two complete DBCS characters).
=MIDB("ä¸ĺ›˝";2;1) returns " " (byte position 2 is not at the beginning of a character in a DBCS string; 1 space character is returned).
=MIDB("ä¸ĺ›˝";2;2) returns " " (byte position 2 points to the last half of the first character in the DBCS string; the 2 bytes asked for therefore constitutes the last half of the first character and the first half of the second character in the string; 2 space characters are therefore returned).
=MIDB("ä¸ĺ›˝";2;3) returns " ĺ›˝" (byte position 2 is not at the beginning of a character in a DBCS string; a space character is returned for byte position 2).
=MIDB("ä¸ĺ›˝";3;1) returns " " (byte position 3 is at the beginning of a character in a DBCS string, but 1 byte is only half a DBCS character and a space character is therefore returned instead).
=MIDB("ä¸ĺ›˝";3;2) returns "ĺ›˝" (byte position 3 is at the beginning of a character in a DBCS string, and 2 bytes constitute one DBCS character).
=MIDB("office";2;3) returns "ffi" (byte position 2 is at the beginning of a character in a non-DBCS string, and 3 bytes of a non-DBCS string constitute 3 characters).