Need urget help with Code... HELP!!!

[Our2CuteKids]

Hi all,

First of all, I know absolutely nothing about PHP. I hired someone to create some pages for me that are basically a selection form that looks at contents of multiple directories, and then based on the users selection, creates an output page with links to each file fitting the criteria. Here's the problem.... the guy has created the scripts using PHP and apparently he is using PHP 4, MySQL 4 and Apache. However, I am running Win2K, IIS, PHP5 and MySQL5.

At first the search page was giving me weird "notices" which he said he wasn't getting on Apache.

Notice: Use of undefined constant group_code - assumed 'group_code' in D:\Inetpub\Data\selection\main.php on line 76

After digging around myself, I found that "group_code" didn't have any single quotes around it. He indicated that it shouldn't need quotes. Once I put in the quotes myself, that error went away.

So on the results page, I am getting an error when there is one (and only one) type of report selected.

The error is:

Notice: Undefined variable: data_file in D:\Inetpub\Data\selection\search.php on line 67

The coder claims that it is due to a conflict with PHP and IIS, basically that I haven't configured it properly. That he's not getting the error on Apache.

Since I was able to figure out that the quotes were needed on the 1st error, I feel that there is an actual error in the code and not PHP/IIS.

Could someone help me figure out what the problem might be. I have questioned him multiple times and am getting the same answer, but since I am clueless about PHP, I don't know what to do.

Thanks in advance for your help!

[ahundiak]

PHP gives the operator some control over what to do when it encounters problems. There are three categories: Errors, Warnings and Notices. Code which generates Notices (as your code does) is technically ok but could have a problem.

The reason your "developer" is not getting these notices is that he had notice reporting turned off. Quite honestly, any decent developer, especially one getting paid, show know about notices.

Add the line
error_reporting(0);
At the top of the main script file and the notices will go away. The script itself may or may not work but you can try it. You can also suppress the notices by editing php.ini. The manual has more information:

http://us3.php.net/manual/en/function.error-reporting.php

And unless there is some sort of gross communication problem I would suggest getting a new developer. If he(or she) cannot deal with this trivial support issue then there is no way that they could handle something more complicated.

[Our2CuteKids]

Thank you so much! That worked perfectly!

And believe me, next project I have... I will not be using this developer.

Thanks again!!! Now I can relax this weekend. 🙂

[mmonaco]

To actually handle this notice you might want to put in a line like
if (!isset($data_file) { do this } else { do this };

[mrgrammar]

ahundiak wrote:

PHP gives the operator some control over what to do when it encounters problems. There are three categories: Errors, Warnings and Notices. Code which generates Notices (as your code does) is technically ok but could have a problem...

Add the line
error_reporting(0);
At the top of the main script file and the notices will go away...

I had error codes turned on for our server and got undefined constant errors all over the place on 2 of my sites. Being a newbie, I didn't know what that meant, but I went searching and was able to figure out and remedy most of the errors. (I'm learning!) I have several questions, though, all related to this same undefined constant error.

1. My guess is that error_reporting(0); turns off all error reporting. (I'm headed off to PHP.net to find out.) Is it possible to turn off just the notices?

2. Why are those single quotes so darn important? (A link to a resource is fine here. I don't mind doing the reading. 🙂 )

3. Not all forms are required (and do not need to be); therefore, when someone submits a form on the site and does not fill in all the fields of the form, my processing script gives me an undefined constant error when it encounters $_POST[fieldname] because no data has been submitted for that field. Is there a remedy for this?

4. I have a situation with a form. The processing script that converts the form data into an email and sends it is located on the same page as the form. It uses

if (isset($_POST[xxx])){
process form here
}

to recognize if the form has been submitted. $_POST[xxx] is a hidden field in the form and is not isset until the form is submitted. Therefore, if the form has not been submitted, that field is not defined until it has been submitted and I get the undefined constant error when the page is accessed. Is there a remedy for this?

Thanks in advance for your help.

bd

[bpat1434]

1. Yes... look at Error Reporting Manual
   If you look further down, you see this in their code:

   // Report all errors except E_NOTICE
   // This is the default value set in php.ini
   error_reporting(E_ALL ^ E_NOTICE);

2. Single quotes tell the php parser that all content contained within is a literal string, and to take it as such. Not sure what you're referencing here, but any further guidance that's needed, I can help you. Also, read up in the manual on strings: Strings Manual

3. You can do a simple check:

   $_POST['fieldname'] = (!isset($_POST['fieldname']))?'':$_POST['fieldname'];

   That says: $_POST['fieldname'] will hold an empty string if $_POST['fieldname'] is not set (i.e. not submitted); otherwise, it is the submitted value.

4. You can do the same type of check, or even try suppressing the error messages by prepending the "isset" with an "@" symbol.

~Brett

[rincewind456]

1. Yes. error reporting

2. Can't answer that one or find a resource to give.

3. You should check for this in your error handling something like

   if ($_POST['xxx'] == ''){
   //do something 
   }

4. You can put

   $_POST['xxx'] = '';

   before your isset check.