What is the use of "ereg_replace("\n","\\n",$row[$j])" expression?

Question

I found one PHP script for get a database backup, there use this expression to do something, that script show call to undefined function ereg_replace() this error, if i remove this line script is working fine... how to replace this function $row[$j] = ereg_replace("\n","\\n",$row[$j]); to that script working finely,

Can anyone assist me..

what version of PHP have you got? `ereg_replace` was deprecated in PHP 5.3 and removed in v7+ — Professor Abronsius, Jun 25 '20 at 13:00
Take a look at this answer: https://stackoverflow.com/a/3649655/2960971 — IncredibleHat, Jun 25 '20 at 13:02
I assume his question is "how to replace this [depreciated] function [with something that does the same so the script works correctly]". — IncredibleHat, Jun 25 '20 at 13:12

Vidal · Answer 1 · 2020-06-25T13:08:52.537

ereg_replace deprecated and remove from newer versions of php (7 and up).

You will have to update your code to use preg_replace

$row[$j] = preg_replace("#\n#","\\n",$row[$j]);

One of the differences between ereg_replace() and preg_replace() is that the pattern must be enclosed by delimiters: delimiter + pattern + delimiter, in this case we are using # so we don't have to escape / that is the usual that is used.

Valid Delimiters:

 /, #, ~, +, %, @, ! , <, >

Reference from php.net

hope it helps.

What is the use of "ereg_replace("\n","\\n",$row[$j])" expression?

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