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Old 11-03-2002, 04:40 PM   #1
antken
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PHP if statements and variables


hi,


this is driving me up the wall

i am typing to make an if statement that will work out if a certain value is given from a database then it will perform a certain task

for example: if the database says the value is 1 then it changes the html to a different button is loaded if its 2 then an extra button is loaded and so on

every time i issue the statement i get back parse errror on line x
i have even looked it up and i cannot find the answer, here is what i am typing

if($lalala == '1'){
dothis();
}

and it fails every time

any ideas??
 
Old 09-24-2003, 02:27 AM   #2
Saraev
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A debugging trick I use when I can't for the life of me find the problem is to comment out the if($lalala section. Right above it, do something like "echo $lalala" so you can see what value is being compared.

As for the code shown, looks fine to me.
 
Old 09-24-2003, 05:28 AM   #3
Robert0380
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try double quotes. may still fail, but try it anyway.


also, where was $lalala declared. i had that problem because i thought some variables were global but werent
 
Old 09-24-2003, 05:41 AM   #4
nephilim
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Even if $lalala isn't declared, it shouldn't give you a parse error, it will just return false.

Could you post some more code, it could be that the line before the if statement is causing you trouble.
 
Old 09-24-2003, 01:45 PM   #5
then
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Re: PHP if statements and variables

Quote:
Originally posted by antken
i am typing to make an if statement that will work out if a certain value is given from a database then it will perform a certain task

every time i issue the statement i get back parse errror on line x
i have even looked it up and i cannot find the answer, here is what i am typing

if($lalala == '1'){
dothis();
}
What's the output of the sql query? I can't see anything wrong with the if. AFAIK, don't need to use quotes when the value of the variable is numeric.

regards
theN
 
  


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