subject:"\[wdvltalk\] Re\: PHP Question"

[wdvltalk] Re: PHP question

2004-03-06 Thread Bj

- Original Message -
From: "William Stewart"

> I have a variable that I want to define as an include of
> a .txt file.  In other words, I want something like:
>
> $subcontent = "include("text.txt")";

Simplest is just to put the whole variable definition into the include file:



where the contents of vardef.txt is the one line:



But if you only want the include file to contain the value to be assigned,
then you have to get php to open the text file for input, and read a line
from it into the variable:



If vardef.txt contained the one line "Happy Birthday" then $subcontent would
now have that as its value.


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[wdvltalk] Re: PHP question concerning folders

2004-02-19 Thread Sheila Fenelon

William Stewart wrote:
I am trying to dynamically include a file based on what folder the file
that is calling it is in.  For instance:
The file www.whatever.com/abc/index.php would include the file:
www.whatever.com/template/abc.php and the file
www.whatever.com/xyz/index.php would include the file:
www.whatever.com/template/xyz.php 

Does this make sense, oh gurus?  I would appreciate your help!

Sure that makes sense. You want the name of script's parent directory. 
You will use that to build the path to the include file.

Here's test script:



// get the full url to the script
$script_name = $_SERVER['SCRIPT_NAME'];
echo "script_name -> $script_name ";
// split the url into an array of it's parts
$parts = explode('/', $script_name);
// the second to last member of the $parts array is the parent directory
$parent_dir = $parts[count($parts)-2];
echo "parent directory -> $parent_dir ";
?>

There are other ways to do this but this will work. You might want to 
add some error checking and make sure $parts has at least two elements.

Sheila
--
Sheila Fenelon
http://www.shefen.com/
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[wdvltalk] Re: PHP Question

2004-01-13 Thread Bj

- Original Message -
From: "Anitha Paruchuri"

> what suprises me is that all the while it did not
> give any problem until i tried entering a
> password starting with numbers.

Well, look at it this way.  By comparing $password with a number you are
forcing it to evaluate to a numeric value.  The numeric value of "123abc" is
123, which is greater than 16 so gives you an error.  "abc123" evaluates to
0, which is less than 16, so no error.

To prove this to yourself try:

print "Numeric value of $password is " .
  1 * $password . "";

Bj


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[wdvltalk] Re: PHP Question

2004-01-13 Thread Anitha Paruchuri

Thanks so much...Huh sometimes you feel like such a stupid to do these small errors. 
But Thanks everyone it is definately the ')' but what suprises me is that all the 
while it did not give any problem until i tried entering a password starting with 
numbers. I think I must have entered atleast 10 records but it worked fine.  Can 
anyone think of any explanation for this?
Thanks again.
Anitha

>>> [EMAIL PROTECTED] 1/13/2004 1:25:09 PM >>>
Anitha Paruchuri wrote:
> The code where it does this is here-- 
> 
>   if (strlen($cpassword)<5 || strlen($cpassword >16))

A misplaced )
You probably want this:
 if (strlen($cpassword)<5 || strlen($cpassword) >16)

Sheila

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[wdvltalk] Re: PHP Question

2004-01-13 Thread Sheila Fenelon

Anitha Paruchuri wrote:
The code where it does this is here-- 

  if (strlen($cpassword)<5 || strlen($cpassword >16))


A misplaced )
You probably want this:
if (strlen($cpassword)<5 || strlen($cpassword) >16)
Sheila



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[wdvltalk] Re: PHP Question

2004-01-13 Thread john . f . hughes

Complete newbie guess

Should

  if (strlen($cpassword)<5 || strlen($cpassword >16))

not be

  if (strlen($cpassword)<5 || strlen($cpassword)>16)

?

  "Anitha  

  Paruchuri"   To:  [EMAIL PROTECTED]  

  <[EMAIL PROTECTED] cc:   

  .niu.edu>            Subject: [wdvltalk] Re: PHP Question

  13/01/2004 19:10 

  Please respond   

  to wdvltalk  

The code where it does this is here--

  if (strlen($cpassword)<5 || strlen($cpassword >16))
   {
   echo "Your password must be between 5 and 16 characters.
Please go back and try again.";
   exit;
 // check password length is ok
   }

I tried printing the values of strlen before the if statement and it does
give me 6 for the length of 123abc but it also gives me the error message
saying "Your password must be."
I now realize that this is not the strlen problem. But What else could be
going wrong here?. I am checking that the user password is between 5 and 16
characters in length. In this example if it is 6 it shouldn't be printing
the message, but it does.

Anitha

>>> [EMAIL PROTECTED] 1/13/2004 11:50:16 AM >>>
Anitha Paruchuri wrote:
> Hi,
> I am trying to check the length of the password entered by the user
> using strlen($password). It works fine when say I enter a password as
> 'abc123' but if I entered '123abc' it somehow doesn't work fine. Am I
> missing something here? Should the $password or the variable for the
> function strlen() cannot start with a number??
>

Strings can be all digits. Do you have a sample of the code that doesn't
work? I've tested this and can't recreate the problem. Here's my test
script:
http://demo.fenelon.us/testCode/strlen_test.php?test3=43434

Change the value for test3 to try different strings.

Sheila

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This information is confidential and may be privi

[wdvltalk] Re: PHP Question

2004-01-13 Thread Anitha Paruchuri

The code where it does this is here-- 

  if (strlen($cpassword)<5 || strlen($cpassword >16))
   {
  echo "Your password must be between 5 and 16 characters. Please go back 
and try again.";
  exit;
// check password length is ok
   }

I tried printing the values of strlen before the if statement and it does give me 6 
for the length of 123abc but it also gives me the error message saying "Your password 
must be."
I now realize that this is not the strlen problem. But What else could be going wrong 
here?. I am checking that the user password is between 5 and 16 characters in length. 
In this example if it is 6 it shouldn't be printing the message, but it does.

Anitha

>>> [EMAIL PROTECTED] 1/13/2004 11:50:16 AM >>>
Anitha Paruchuri wrote:
> Hi,
> I am trying to check the length of the password entered by the user
> using strlen($password). It works fine when say I enter a password as
> 'abc123' but if I entered '123abc' it somehow doesn't work fine. Am I
> missing something here? Should the $password or the variable for the
> function strlen() cannot start with a number??
> 

Strings can be all digits. Do you have a sample of the code that doesn't 
work? I've tested this and can't recreate the problem. Here's my test 
script:
http://demo.fenelon.us/testCode/strlen_test.php?test3=43434 

Change the value for test3 to try different strings.

Sheila

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[wdvltalk] Re: PHP Question

2004-01-13 Thread Bj

- Original Message -
From: "Anitha Paruchuri"

> I am trying to check the length of the password entered
> by the user using strlen($password). It works fine when
> say I enter a password as 'abc123' but if I entered
> '123abc' it somehow doesn't work fine. Am I missing
> something here?

Either string should be treated the same - try putting in the code:

print "Length of abc123: ";

...then the same for 123abc, and you should see them both print on the page
as 6.  If you are getting a different result when dealing with form fields,
then something is going wrong in the process.

First try using $HTTP_POST_VARS["password"] instead of $password, to make
sure you are actually looking at the value passed from the form.


Bj


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[wdvltalk] Re: PHP Question

2004-01-13 Thread Sheila Fenelon

Anitha Paruchuri wrote:
Hi,
I am trying to check the length of the password entered by the user
using strlen($password). It works fine when say I enter a password as
'abc123' but if I entered '123abc' it somehow doesn't work fine. Am I
missing something here? Should the $password or the variable for the
function strlen() cannot start with a number??
Strings can be all digits. Do you have a sample of the code that doesn't 
work? I've tested this and can't recreate the problem. Here's my test 
script:
http://demo.fenelon.us/testCode/strlen_test.php?test3=43434

Change the value for test3 to try different strings.

Sheila

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