CODEWITHSUNDEEP
phparrays
Djacksway
Djacksway

Reputation: 457

append csv to php array

I can not get this PHP array to return any values for me. The CSV sheet consists of 10 numbers. The only output I get is "array"

$data = array();
if (($handle = fopen("fullbox.csv", "r")) !== FALSE) {
    while(($row = fgetcsv($handle, 1000, ",")) !== FALSE) {
        $data[] = $row;
    }
}

echo $data[1];

Any help would be greatly appreciated.

Upvotes: 1

Views: 157

Answers (3)

Michael Berkowski
Michael Berkowski

Reputation: 270677

Your code is already correct.

$data is multidimensional array, each element is an array itself. When you echo $data[1] you are asking PHP to write a string representation of a more complex array variable, which PHP handles by outputting array instead of its contents.

Instead, var_dump() it to see what it contains.

var_dump($data);

A single value would be accessed via something like :

echo $data[1][0];

Edit after comment:

If the CSV contains only one value per row, access it directly via $row[0] when appending to the output array in order to get it as a 1D array:

if (($handle = fopen("fullbox.csv", "r")) !== FALSE) {
  while(($row = fgetcsv($handle, 1000, ",")) !== FALSE) {
    $data[] = $row[0];
  }
}
// $data is now a 1D array.
var_dump($data);

Upvotes: 3

GhostInTheSecureShell
GhostInTheSecureShell

Reputation: 1010

$data[1] is the second element in your $data variable ($data[0] would be the first).

You are getting an array back instead of a value which is why PHP is echoing the word array.

This should work better.

$data[0][0]

Your $data variable is a multidimensional array and you are only going one level deep.

Upvotes: 0

hellsgate
hellsgate

Reputation: 6005

Thats because $data[1] is an array of the 10 numbers in the CSV row, so $data is now a multi-dimensional array. echo $data[1][0] will output one of the numbers

Upvotes: 0

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