So you can't say print('what up doc $mycarriagereturnvariable'); and expect it to output the value of $mycarriagereturnvariable. Next, you *can* use this construct within *double* quotes. Finally, there are special characters - "Escaped characters" you can use with PHP to generate a carriage return :
These include :
\t : Tab character \r : Carriage return character \n : Newline character
So your string should be :
print("what up doc\r\n");
Hope that helps Neil.
At 16:23 16/04/2004 +0000, you wrote:
MIME-Version: 1.0 Content-Type: text/plain; charset="iso-8859-1" Date: Fri, 16 Apr 2004 11:23:51 -0500 Message-ID: <AED295B4597E884A82081E567887C4F30E1C90@xxxxxxxxxxxxxxxxxxxxx> From: "Hull, Douglas D" <ddhull@xxxxxx> To: "Note To php mysql List (E-mail)" <php-db@xxxxxxxxxxxxx> Subject: putting strings together with a linefeed
For example, say I have a variable $mytext = 'what up doc'. Now I want to add the text 'not much' to the end of this variable. But I want to place a return or linefeed after the original text so when I echo $mytext it looks like:
what up doc not much
I tried: $mytext = 'what up doc'; $mytext .= chr(10) . 'not much'; I also tried chr(13)
10 is linefeed and 13 is carriage return.
Thanks for your previous help as well. Doug
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