Re: fgetcsv doesn't return an array?

On Thu, 15 Mar 2012 12:09:53 -0500, Jay Blanchard <jay.blanchard@sigmaphinothing.org> sent:

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 |

Just want to clarify something not directly related to the OP's question: This:

153 |$csvCurrentLine = array();

Does not set a *type* for $csvCurrentLine, so if you're next line is this:

154 $csvCurrentLine = fgetcsv($csvFile, 4096, ',');

Any previous assignments are lost; the notion that presetting $csvCurrentLine to an array has no bearing when you then immediately set the same variable to something else. The "type" $csvCurrentLine has after line 154 is entirely dependent on what fgetcsv() is returning, and nothing that happened prior to line 154. -- Tamara Temple aka tamouse__ May you never see a stranger's face in the mirror