how to remove new lines and returns from php string?

Question

A php variable contains the following string:

text
text2

item1
item2

        

Answer 1

You have to wrap \n or \r in "", not ''. When using single quotes escape sequences will not be interpreted (except \' and \\).

The manual states:

If the string is enclosed in double-quotes ("), PHP will interpret more escape sequences for special characters:

• \n linefeed (LF or 0x0A (10) in ASCII)

• \r carriage return (CR or 0x0D (13) in ASCII)\

• (...)

Answer 2

Something a bit more functional (easy to use anywhere):

function replace_carriage_return($replace, $string)
{
    return str_replace(array("\n\r", "\n", "\r"), $replace, $string);
}

Using PHP_EOL as the search replacement parameter is also a good idea! Kudos.

Answer 3

$str = "Hello World!\n\n";
echo chop($str);

output : Hello World!

Answer 4

Correct output:

'{"data":[{"id":"1","reason":"hello\\nworld"},{"id":"2","reason":"it\\nworks"}]}'

function json_entities( $data = null )
{           
    //stripslashes
    return str_replace( '\n',"\\"."\\n",
        htmlentities(
            utf8_encode( json_encode( $data)  ) , 
            ENT_QUOTES | ENT_IGNORE, 'UTF-8' 
        )
    );
}

Answer 5

This should be like

str_replace("\n", '', $str);
str_replace("\r", '', $str);
str_replace("\r\n", '', $str);

Answer 6

You need to place the \n in double quotes.
Inside single quotes it is treated as 2 characters '\' followed by 'n'

You need:

$str = str_replace("\n", '', $str);

A better alternative is to use PHP_EOL as:

$str = str_replace(PHP_EOL, '', $str);