Re: fgetcsv doesn't return an array?

On Thu, Mar 15, 2012 at 1:29 PM, Jay Blanchard <jay.blanchard@sigmaphinothing.org> wrote: > On 3/15/2012 12:23 PM, Matijn Woudt wrote: >> >> On Thu, Mar 15, 2012 at 6:09 PM, Jay Blanchard >> <jay.blanchard@sigmaphinothing.org>  wrote: >>> >>> I thought that fgetcsv returned an array. I can work with it like an >>> array >>> but I get the following warning when using it >>> >>> |Warning:  implode(): Invalid arguments passed on line 155 >>> >>> 154     $csvCurrentLine = fgetcsv($csvFile, 4096, ','); >>> 155     $currentLine = implode(",", $csvCurrentLine); >>> | >>> >>> Everything that I have read online indicates that it is because >>> $csvCurrentLine is not declared as an array. Adding a line to the code >>> >>> 153 |$csvCurrentLine = array(); >>> >>> Does not get rid of the warning. The warning isn't interfering with the >>> script working, but it us annoying. Does anyone know the solution? >>> >>> Thanks! >>> >>> Jay >>> | >> >> Are you using this in a loop or something? fgetcsv returns FALSE on >> errors, including End Of File. >> >> [/snip] > > I am using it in a loop. End Of File is an error? Yes. "fgetcsv() returns NULL if an invalid handle is supplied or FALSE on other errors, including end of file." I usually use it in a while loop like this: <?php while ($row = fgetcsv($csvFilePointer)) { // ... process the row } ?> Andrew