Re: fgetcsv doesn't return an array?

On Thu, Mar 15, 2012 at 6:40 PM, Andrew Ballard <aballard@gmail.com> wrote: > 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 > } > > ?> Please, don't do that. Use this instead: while (($csvCurrentLine = fgetcsv($csvFile, 4096, ',')) !== FALSE) { }