| Re: Undefined variable [message #181237 is a reply to message #181235] |
Wed, 24 April 2013 03:08   |
Richard Yates
Messages: 86
Registered: September 2013
Karma:
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On Tue, 23 Apr 2013 19:16:54 -0700 (PDT), Question Boy
<question(dot)boy(at)hotmail(dot)com> wrote:
> I have an simple MySQL/PHP app and it appears to be functional but the
> webmaster has informed me that it is throwing lots of errors. So he
> showed me the log file and I am trying to remedy the issues, but have
> a question.
>
> For instance, I have a block of code, such as:
>
> if ($iBookedBy=="Other") {
> echo '<option selected value="Other">Other</option>';
> } else {
> echo '<option value="Other">Other</option>';
> }
>
> which the log file reports as:
>
> PHP Notice: Undefined variable: iBookedBy
>
>
>
>
> Now I thought of trying:
>
> if (isset($iBookedBy)==TRUE && $iBookedBy=="Other") {
> echo '<option selected value="Other">Other</option>';
> } else {
> echo '<option value="Other">Other</option>';
> }
>
>
> but am not sure if the server will break because the variable isn't
> set or if it will still throw an error because of the second,
> original, part of the if statement? Is this a good way to handle the
> problem, or am I going about this the wrong way and there is a better
> method?
>
> Thank you for your help.
>
> QuestionBoy
1. Yes, it will prevent the error notice.
2. You do not need the ==TRUE. isset($iBookedBy) means the same
isset($iBookedBy)==TRUE.
3. Your asking the question suggests that you are trying to write code
and run it directly on a production server rather than testing on a
local server. The webmaster will be happier, and you will be much more
productive, if you set up a local server like WAMP.
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