Question

I am doing something like:

$outputFile = getCurrentDBSnapshot($data);

where $data is the resource stream that am passing in, basically from command prompt am passing an file and am opening it using fopen for writing with ‘w+’ permissions, now getCurrentDBSnapshot would get the current state of a table and would update the $data csv file, so basically $outputFile would be updated with the current state of database table, now I want to var_dump or print the value of $outputFile to see the data present into it.

But when I do

$this->fout = fopen($outputFile,'r') or die('Cannot open file');
$test = fgetcsv($outputFile,5000,";");
var_dump($test);

It gives me an error saying that it expects parameter 1 to be a string type and am passing resource.

My goal to see the contains of $outputFile

and so my question is that

How can I see the contains present in $outputFile or how can I see what getcurrentDBSnapshot function is returning me ?

Answer 1

fegtcsv takes as first parameter a file handle, not a filename. You’ll need to do something like:

$this->fout = fopen($outputFile,'r') or die('Cannot open file');
while ($test = fgetcsv($this->fout,5000,";"))
{
    var_dump($test);
}

Note that fgetcsv merely gets a single line of the file, analogously to fgets.

Also I’m not sure why you’re passing a semicolon as the third argument to fgetcsv. CSV stands for comma-separated-value; are you sure your file is semicolon-delimited?