Question: I want to extract the part before the 1st `/`, how can I do that?

```
 SELECT REGEXP_SUBSTR('prAct1-03/cPMI-01/iPEPC9-01/t35s-08','(pr.*)/',1,1) as promoter 
   FROM dual;

```

did not work, and seems oracle did not support Lookahead and Lookbehind.
Thanks!

Answer: 
I would just use the base string functions here:

```
SELECT SUBSTR(col, 1, INSTR(col, '/') - 1) AS first_part
FROM yourTable;

```

[Demo
----](https://dbfiddle.uk/?rdbms=oracle_11.2&fiddle=9f93f9b1b0629326b6be1ff700bcc170)

If you really wanted to use regex, then try:

```
SELECT REGXP_SUBSTR(col, '^[^/]+')
FROM yourTable;

```
One option uses `REGEXP_REPLACE()` function :

```
SELECT REGEXP_REPLACE('prAct1-03/cPMI-01/iPEPC9-01/t35s-08','(\S)(/.*)','\1') AS promoter
  FROM dual;

```

`[Demo](https://dbfiddle.uk/?rdbms=oracle_11.2&fiddle=95d17d21eca9223566e078a0e00bb0b3)`