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eyn
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Joined: 11 Sep 2004
Posts: 32
Location: Canada
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 Posted: Sun Jan 02, 2005 8:43 am Post subject: PHP newbie question |
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Okay, I'm learning PHP using some eBook, I have some newbie question here. How to use a variable without getting the "undefined variable" notice from the compiler? For example, I have this line:
| Code: | | $movie_list=array_merge($movie_list, $movie); |
This is the first line where this variable is used. I want $movie_list to be an empty array, so I can append my $movie array to it each time the "foreach" loop goes in order form a movie list.
Am I suppose to declare an empty array such that $movie_list is not undefined when the compiler see it? How to declare an empty array?
Another newbie question, when you use print function such as below:
| Code: | | print "<SPAN class='headline'>My title</SPAN>"; |
I am forced to use ' instead of " for my headline css class, or else the compiler will give me error. Can I use "headline" instead of 'headline' without getting an error?
I hope my question make sense. I'm using PHP 4.3, testing on Abyss Web Server. Ah, also where is the php.ini file, and how do I use it? Thank you very much! |
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eyn
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Joined: 11 Sep 2004
Posts: 32
Location: Canada
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| Posted: Sun Jan 02, 2005 8:53 am Post subject: Well, guess what? |
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Ha, I used the following code and it works!
| Code: | | $movie_list = array(); |
Oh well, I'm not really sure if this is the way you should do it, any idea?
I still have the other question I posted though. :p |
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TRUSTAbyss
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Joined: 29 Oct 2003
Posts: 3752
Location: USA, GA
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| Posted: Sun Jan 02, 2005 9:42 am Post subject: |
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Im not too sure about your first question because im also still learning a few
things about PHP , here's an example as to why you have to use 'headline'
Since " is part of a print statement or a value set for a variable , you have
to escape the " in headline so PHP doesn't interpret it and show an error.
Example:
To tell PHP not to interpret the " in headline , just add a \ in the back of " so
just use \"headline\" for every quote and this will tell PHP not to interpret it. |
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