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 );
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( $columns);
As of your code now, you are not returning a result from your MySQL server. For that, you need to pass your mysql_query result to a mysql_fetch_*
function.
For your use, you can do mysql_fetch_assoc($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;