Variable not recognised in IF statement

tonyrhills

I am new to this forum and relatively new to PHP. I have been developing a web site to be used by individual charities and come across a small problem where I wonder if anyone can help.

The web site is designed so that charities recommend the site to their supporters using a unique URL which I allocate and creates their landing page. Their name is coded in as $cname=’charity name’. This then carries throughout the site and creates a bespoke feel to the viewer.

If the name is not present the default for $cname is ‘your chosen charity’ and this works throughout the text in the same way as a bespoke name.

All works well until a page I created with a form to be completed. In this form is a field called $this_cname for the charity’s name, which is pre-populated. However, if the charity name is not available, I want the field left blank and not pre-populated with ‘your chosen charity’.

At the start of the coding for this page, I do the usual $cname = $_POST['cname']; and then

if ($cname == "your chosen charity") {$this_cname = “”;}
else {$this_cname = $cname;}

Then, to see how it is working (although this will not be in the final version):

echo '' . $this_cname . '';
echo '' . $cname . '';

This coding does not work. Whatever $cname is, it is echoed in both results and the field remains pre-populated with ‘your chosen charity’ if that is what $cname is.

Researching the web I discovered a modified wording:

if ($_POST['cname'] == "your chosen charity") {$this_cname = “”;}
else {$this_cname = $cname;}

This also does not work. $this-cname remains the same as $cname.

However, if I delete the $cname = $_POST['cname']; and substitute $cname=”your chosen charity” the code works and the field remains blank, or fills in if I change $cname to anything else.

Clearly, the $_POST is not being recognised by the IF statement but is by the echo statement.

I am at a loss to understand where I am going wrong.

Any help would be most appreciated.

Tony Hills

laserlight

I suggest:

if (isset($_POST['cname']) && $_POST['cname'] !== 'your chosen charity')
{
    $this_cname = $_POST['cname'];
}
else
{
    $this_cname = '';
}

You should also check your form: are you sure the field name is 'cname'?

tonyrhills

laserlight;11042371 wrote:
I suggest:
if (isset($_POST['cname']) && $_POST['cname'] !== 'your chosen charity')
{
    $this_cname = $_POST['cname'];
}
else
{
    $this_cname = '';
}
You should also check your form: are you sure the field name is 'cname'?

Hi Laserlight

Thanks for your quick reply. I have tried this coding but it still gives the wrong result. So I have tried some more experimenting and found the following:

Using the following coding:

$cname = $_POST['cname'];
echo '' . $cname . '';
$this_cname = $cname;
echo '' . $this_cname . '';
if ($cname == ’your chosen charity’) {$this_cname = 'test';}
echo '' . $this_cname . '';
echo '' . $cname . '';

when $cname is the default (your chosen charity) the four ECHOs return:

your chosen charity
your chosen charity
your chosen charity
your chosen charity

If $cname is the name of a charity (e.g. RSPCA) the four ECHOs return:

RSPCA
RSPCA
RSPCA
RSPCA

However, if I change the default to a numeric (e.g. 1234) and change the IF statement to

if ($cname == 1234) {$this_cname = 'test';}

when $cname is the default (1234) the four ECHOs return:

1234
1234
Test
1234

If $cname is the name of a charity (e.g. RSPCA) the four ECHOs return:

RSPCA
RSPCA
RSPCA
RSPCA

Exactly as I expect it to work. So the IF statement appears not to be responding to letters but will to a numeric value. Has anyone experienced this? Is it a fault in my coding or in the PHP set up on the server?

Tony

Derokorian

Yout might want to check for extra whitespace on the incoming string using [man]trim[/man]. Something like:

if (isset($_POST['cname']) && trim($_POST['cname']) !== 'your chosen charity')
{
    $this_cname = $_POST['cname'];
}
else
{
    $this_cname = '';
}

laserlight

tonyrhills wrote:
I have tried this coding but it still gives the wrong result.

I can tell you that my code is correct, so if "it still gives the wrong result", then the problem probably lies elsewhere in your code.

Post the smallest and simplest PHP script that demonstrates the wrong result. Show us the form.

tonyrhills wrote:
Using the following coding:

$cname = $_POST['cname'];
echo '' . $cname . '';
$this_cname = $cname;
echo '' . $this_cname . '';
if ($cname == ’your chosen charity&#8217😉 {$this_cname = 'test';}
echo '' . $this_cname . '';
echo '' . $cname . '';

when $cname is the default (your chosen charity) the four ECHOs return:

your chosen charity
your chosen charity
your chosen charity
your chosen charity

You don't need to prepend and append empty strings to $cname. In line with Derokorian's suggestion of the use of trim, I suggest having visible delimiters instead so that you can see if there's any whitespace before or after the string:

if (isset($_POST['cname']) && trim($_POST['cname']) !== 'your chosen charity')
{
    $cname = $_POST['cname'];
}
else
{
    $cname = 'test';
}

echo '"' . $cname . '"';

If you have $cname, $this_cname is unnecessary and just causes confusion. Pick a single variable name. Unconditionally assigning $cname to $this_cname is just a waste of effort.

tonyrhills wrote:
However, if I change the default to a numeric (e.g. 1234) and change the IF statement to

Notice the !== in my code: I am playing it safe in terms of type.