Storing values of multidimensional array into variables

Question

I have the following code:

$res = $db->getArticles($db);
$hey = print_r($res, TRUE);
echo $hey['0'];

Here is the code for $db->getArticles:

public function getArticles() {
    $array = array();
    try {
        $sth = $this->db->prepare("SELECT * FROM posts ORDER BY id");
        $sth->execute();
        foreach ($sth->fetchAll(PDO::FETCH_ASSOC) as $row) {
            $array[] = $row;
        }
        return $array;
    } catch (Exception $e) {

    }
}

The logic is that the function queries the database and puts everything into an array. The number of rows in the table will always be pretty low, so performance isn't an issue.

The following is the output of the first code snippet above:

echo $hey['0']; // prints "A" (without the quotes).
echo $hey['1']; // prints "r"
echo $hey['2']; // prints "r"
echo $hey['3']; // prints "a"
echo $hey['4']; // prints "y"

As you can see, it spells out the word "Array."

echo $hey prints the following:

Array ( [0] => Array ( [id] => 1 [author] => 0 [content] => This Is a Test [publication_date] => 1380380992 ) )

My end goal is to store each individual value into a variable and do something with it. This would all happen inside a for loop which runs on the array so I can get information from each row.

What am I doing wrong?

print_r just prints a string detailing the array, the true option just returns the string instead of printing it out directly. It doesn't put the actual data array into the variable — Patrick Evans
– Patrick Evans, Commented Sep 30, 2013 at 14:50

Nick Rolando · Accepted Answer · 2013-09-30 15:00:10Z

1

$hey = print_r($res, TRUE);

This will return a string that gives the info of the array $res. If you echo it, you should expect to see the same as what print_r($res); displays, as you've shown. $res is already an array of your query data, so loop over that.

foreach($res as $row) {  //row loop
  foreach($row as $col) {  //column loop
    //do something
  }
}

edited Sep 30, 2013 at 15:00

answered Sep 30, 2013 at 14:53

Nick Rolando

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2 Comments

DemCodeLines Over a year ago

Tried replacing // do something with echo $res[$row][$col]; and got "Warning: Illegal offset type" on that echo line 4 times.

Nick Rolando Over a year ago

The variable would be just $col. If you did a nested for loop, the syntax would be like you tried.

Joran Den Houting · Accepted Answer · 2013-09-30 14:54:31Z

0

$array = array( array( ID => 1, 
                  author => 0,
                  content => "This Is a Test",
          publication_date => 1380380992
                )
         );



echo $array['0']['author'];

This works for me...

just echo echo $array['0']['author']; and replace author by the field you need.

edited Sep 30, 2013 at 14:54

answered Sep 30, 2013 at 14:45

Joran Den Houting

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5 Comments

DemCodeLines Over a year ago

Hey, thanks for the quick reply! I tried it but its still the same, except I have a notice too now: Warning: Illegal string offset 'content' in...A. The error is on that line.

Joran Den Houting Over a year ago

Can you maybe try again?

DemCodeLines Over a year ago

echo $hey['0']['0']; prints just "A" without the warning, but I get "Notice: Uninitialized string offset: 1" if I do echo $hey['0']['1']; or echo $hey['0']['2'];

Patrick Evans Over a year ago

the problem is print_r does not return array data, using $hey['0'] or any other array syntax with $hey you are just accessing the individual characters that make up the returned string.

Joran Den Houting Over a year ago

You need to use the array, not $hey ;) Look at my answer.

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