object to array

Question

I need to get an object $resultInput to be viewed as an array (like $resultInput commented out). However when I use the function object to array I get an error Resource id #5. What is the issue?

require('dbc.php');

mysql_select_db($db);
$resultInput = mysql_query("SHOW COLUMNS FROM about WHERE Field NOT IN ('id', 'created', 'date_modified', 'last_modified', 'update', 'type', 'bodytext') AND Field NOT REGEXP '_image'"); // selects only the columns I want

//$resultInput = array('page_header', 'sub_header', 'content', 'content_short');

function objectToArray($object){
    if(!is_object($object) && !is_array($object)){
        return $object;
    }
    if(is_object($object)){
        $object = get_object_vars( $object );
    }
    return array_map( 'objectToArray', $object );
}

// convert object to array
$array = objectToArray( $resultInput );

//show the array
print_r( $array );

Have you checked the manual?php.net/manual/en/function.mysql-query.php — xdazz
– xdazz, Commented Jul 2, 2012 at 14:28
It's better not to use mysql_* at all. mysql_* is getting deprecated. Use MySQLi or PDO instead. php.net/mysql_connect — Sherlock
– Sherlock, Commented Jul 2, 2012 at 14:30

nickb · Accepted Answer · 2012-07-02 15:30:08Z

5

You are getting this error because mysql_query() returns a resource, not an object. To get an object, you would need to use one of the mysql functions to retrieve the rowset as an object (i.e. mysql_fetch_object()). However, you don't need the objectToArray function, as you can just use mysql_fetch_array to fetch the rowset as an array, like so:

$result = mysql_query("...");
$resultInput = mysql_fetch_array( $result);

Also, I'm sure somebody will comment about using the now-deprecated mysql functions. But, just to be thorough, you should be looking into using either mysqli or PDO.

So you got this as output:

array(12) {
    [0] = > string(11)"page_header"
      ["Field"] = > string(11)"page_header" 
    [1] = > string(11)"varchar(50)" 
      ["Type"] = > string(11)"varchar(50)" 
    [2] = > string(3)"YES" 
      ["Null"] = > string(3)"YES" 
    [3] = > string(0)"" 
      ["Key"] = > string(0)"" 
    [4] = > NULL
      ["Default"] = > NULL
    [5] = > string(0)"" 
      ["Extra"] = > string(0)""
}

Which is describing the page_header column. To just display the column names for all columns, you would do:

$result = mysql_query("..."); $columns = array();
while( $row = mysql_fetch_array( $result) {
    $columns[] = $row['Field'];
}

var_dump( $columns);

edited Jul 2, 2012 at 15:30

answered Jul 2, 2012 at 14:28

nickb

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3 Comments

Alex Over a year ago

I dont want to fetch the rowset I want to get the column names as defined in SHOW COLUMNS from. Which I realize is mysql_fetch_assoc

nickb Over a year ago

... The column names will be returned in the rowset. Do var_dump( $resultInput); and you'll see them...

Alex Over a year ago

Interesting... forgive me for I am a novice. But how does one get just the names of the columns into an array instead of this: array(12) { [0]=> string(11) "page_header" ["Field"]=> string(11) "page_header" [1]=> string(11) "varchar(50)" ["Type"]=> string(11) "varchar(50)" [2]=> string(3) "YES" ["Null"]=> string(3) "YES" [3]=> string(0) "" ["Key"]=> string(0) "" [4]=> NULL ["Default"]=> NULL [5]=> string(0) "" ["Extra"]=> string(0) "" } I dont need all this varchar and key information

Mike Mackintosh · Accepted Answer · 2012-07-02 14:44:31Z

0

As of your code now, you are not returning a result from your MySQL server. For that, you need to pass your mysql_query result to a mysql_fetch_* function.

For your use, you can do mysql_fetch_assoc($result);, which will retain column/value association.

Now, if you were using MySQLi and had mysqli_fetch_object(), you can use the below to turn the object to array, and vice versa.

You can do the following:

Typecasting:

$array = new array();
$object = (object) $array;

or

$object = new stdClass();
$array = (array) $object;

answered Jul 2, 2012 at 14:44

Mike Mackintosh

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