what am I doing wrong

[hougie40]

Im a vb guy trying to figure out some php.

For some reason it wont let me append these together.

$variable = "test";

echo "c:\websites\me nu\" . $variable;

dont understand...pl ease help.

[Dikkie Dik]

ho******@hotmai l.com wrote:

Im a vb guy trying to figure out some php.

For some reason it wont let me append these together.

$variable = "test";

echo "c:\websites\me nu\" . $variable;

dont understand...pl ease help.

Is the second quote seen as escaped by the backslash? Maybe it will echo
"c:\\websites\\ menu\\" . $variable

Best regards

[Aphrael]

ho******@hotmai l.com a écrit :

Im a vb guy trying to figure out some php.

For some reason it wont let me append these together.

$variable = "test";

echo "c:\websites\me nu\" . $variable;

dont understand...pl ease help.

Ho, you do tell him to consider the final " as a text " instead of the
end of the string... (\ is the escape caracter...)

You just need to tell him it is the \ you want to escape... echo
"c:\websites\me nu\\" . $variable;

By the way, isn't variable a reserved word ?

Aphrael...
--
"La demande mondiale d'ordinateurs n'excédera pas cinq machines."
(Thomas Watson, Fondateur d'IBM, 1945)

[Geoff Berrow]

I noticed that Message-ID:
<11************ **********@g43g 2000cwa.googleg roups.com> from
ho******@hotmai l.com contained the following:

$variable = "test";

echo "c:\websites\me nu\" . $variable;

The backslash is the escape character. For it to be a literal backslash
you have to escape it with...uh...a backslash.

echo "c:\\websites\\ menu\\" . $variable;

Or you could do

echo "c:\\websites\\ menu\\$variable ";
--
Geoff Berrow (put thecat out to email)
It's only Usenet, no one dies.
My opinions, not the committee's, mine.
Simple RFDs http://www.ckdog.co.uk/rfdmaker/

[Paul Herber]

On Mon, 19 Sep 2005 17:37:08 +0100, Geoff Berrow
<bl******@ckdog .co.uk> wrote:

I noticed that Message-ID:
<11*********** ***********@g43 g2000cwa.google groups.com> from
ho******@hotma il.com contained the following:

$variable = "test";

echo "c:\websites\me nu\" . $variable;

The backslash is the escape character. For it to be a literal backslash
you have to escape it with...uh...a backslash.

echo "c:\\websites\\ menu\\" . $variable;

or put a bit less stress on the server

echo 'c:\websites\me nu\' . $variable;
--
Regards, Paul Herber, Sandrila Ltd. http://www.pherber.com/
SanDriLa - SDL/MSC/TTCN/UML2 application for Visio http://www.sandrila.co.uk/

[Chung Leong]

Need to escape backslashes in PHP. Try echo "c:\\websites\\ menu\\" .
$variable.

[JDS]

On Mon, 19 Sep 2005 09:20:13 -0700, ho******@hotmai l.com wrote:

For some reason it wont let me append these together.

$variable = "test";

echo "c:\websites\me nu\" . $variable;

dont understand...pl ease help.

Blame it on Microsoft

--
JDS | je*****@example .invalid
| http://www.newtnotes.com
DJMBS | http://newtnotes.com/doctor-jeff-master-brainsurgeon/

[badr]

actullay no need for blaming any 1
the correct way is
- to escape the last backslash:
$var="test";

echo "C:\web\menu\\" .$var;
---------------------------------------------
the optimum solution for the case is :
$var='test'; //singl qoutes
echo 'C:\web\menu\\' .$var; //singl qoutes

---------------------------------------------
in double qutes PHP will seek for some varaiable inluded with a string
so in first solution you may make it like :

$var="test";

echo "C:\web\menu\\$ var";