Why are the commas in my array reading as a string?

Question

This is what I am trying to do:

1 - I have an array with a string.

$photographer_array = "'Alberto Korda', 'Annie Leibowitz', 'Ansel Adams'";

2 - I am trying to populate an array with that string.

array($photographer_array);

What it is doing is creating an array with one single entry including all the commas. Why is it reading the commas as a string? Is there anyway I can make it not read the commas as a string and instead use them to separate the values in the array?

Answer 1

If you've got access to PHP >= 5.3.0, then you can use str_getcsv, which parses a comma-separated string into an array.

Answer 2

Explode() is what you need...

http://php.net/manual/en/function.explode.php

Answer 3

That's because $photographer_array is a string and using array($photographer_array) simply puts your string to the array.

Perhaps you want to do:

$photographer_array = explode(',', "'Alberto Korda', 'Annie Leibowitz', 'Ansel Adams'");

This will give you

array(3) {
  [0]=>
  string(15) "'Alberto Korda'"
  [1]=>
  string(18) " 'Annie Leibowitz'"
  [2]=>
  string(14) " 'Ansel Adams'"
}

Or you want:

$photographer_array = array('Alberto Korda', 'Annie Leibowitz', 'Ansel Adams');
// ...

EDIT:

Regarding the code you posted - the following will give you an array of the $photographer_row['photographer_name']:

// ...
$photographer_array = array();
while($photographer_row = $result->fetch_assoc()) {
    $photographer_array[] = $photographer_row['photographer_name'];
}

Answer 4

Im actually creating an array using values from a query just so you know why I am doing it this way. This is the code I am using to create the string for use in my array.

$sql = "SELECT photographers.photographer_name
     FROM photographers
     ORDER BY photographer_name ASC
     LIMIT 3";

$result = $conn->query($sql) or die(mysqli_error());

$photographer_array = '';

while($photographer_row = $result->fetch_assoc()) {
$photographer_array .= "'" . $photographer_row['photographer_name'] . "', ";
}

$photographer_array = rtrim($photographer_array, ", ");

array($photographer_array);

Answer 5

Sorry to say that, but you cannot populate an array this way, since strings are usually not parsed and executed. All you can do is:

$photographer_array = array('Alberto Korda', 'Annie Leibowitz', 'Ansel Adams');

Or this (EDIT: Adapted it to your query code):

$photographer_array = array();

while ($photographer_row = $result->fetch_assoc()) {
    $photographer_array[] = $photographer_row['photographer_name'];
}

P.S.: You just love long variable names, don't you? ;)

Answer 6

It's reading the string as a string becuase you enclosed it in quotation marks ( " ). If you want to do it right, try:

$array = array('Alberto Korda', 'Annie Leibowitz', 'Ansel Adams');

Or:

$string = "Alberto Korda, Annie Leibowitz, Ansel Adams"; $arr = explode(',', $string);

Answer 7

you'll notice that the variable $photographer_array is set to a single string (the value of which is everything between the double quotes). The array function in php accepts a list of objects delimited by commas. While there are commas in your $photographer_array variable, they are considered to be characters in the string rather than actual delimiters, so you're really just creating an array with only one value which happens to be a string. What you probably want is the following:

$photographer_array = array('Alberto Korda', 'Annie Leibowitz', 'Ansel Adams');

Answer 8

If you need a string to make a array use:

$photographer_array = "'Alberto Korda', 'Annie Leibowitz', 'Ansel Adams'";

eval("\$myArray = array(" . $photographer_array . ");");

And after use $myArray