PHP: mysql_fetch_object

Описание

object mysql_fetch_object ( resource result )

Возвращает объект со свойствами, соответствующими колонкам в обработанном ряду или FALSE, если рядов больше нет.

mysql_fetch_object() работает аналогично mysql_fetch_array(), с единственным отличием -- функция возвращает объект, вместо массива. Это, кроме всего прочего, означает, что вы сможете работать с полями только по имени колонок. Числа не могут быть свойствами объекта.

Замечание: Имена полей, возвращаемые этой функцией, регистро-зависимы.

<?php /* корректно */ echo $row->field; /* не корректно */ echo $row->0; ?>

В плане скорости эта функция аналогична mysql_fetch_array() и почти также быстра, как mysql_fetch_row() (разница незначительна).
Пример 1. Пример использования mysql_fetch_object()

<?php mysql_connect("hostname", "user", "password"); mysql_select_db("mydb"); $result = mysql_query("select * from mytable"); while ($row = mysql_fetch_object($result)) { echo $row->user_id; echo $row->fullname; } mysql_free_result($result); ?>

См. также mysql_fetch_array(), mysql_fetch_assoc() и mysql_fetch_row().

mysql_fetch_object

trithaithus at tibiahumor dot net
21-Aug-2005 01:56


This method offers a nice way to fetch objects from databases. As Federico at Pomi dot net mentioned it doesn't work native as the type of the object fetched isn't the right one, but with a small typecast it works flawlessly. 



<?php

function ClassTypeCast(&$obj,$class_type){

   if(class_exists($class_type)){

       $obj = unserialize(preg_replace("/^O:[0-9]+:\\"[^"]+\\":/i",

         "O:".strlen($class_type).":"".$class_type."\\":", serialize($obj)));

   }

}



class Foo

{

    var $foo;

    var $bar;

    function get_from_db()

    {

        mysql_connect();

        mysql_select_db();

        $res = mysql_query("SELECT foo,bar from my_table");

        

        $fetched_object = mysql_fetch_object($res);

        ClassTypeCast($fetched_object,"Foo");

        $this = $fetched_object;

    }

}

?>

backglancer at hotmail
07-Aug-2005 03:07


Watch out for mysql_fetch_object() to return all values as strings.

if you try to do

<?

$p = mysql_fetch_object($some_sql);

// and then try to do something like



$money = $p->dollars + $p->cents;

?>



You may experience "Unsupported operand types"



so always cast them both as (int) 's!!

q
09-Jul-2004 07:31


Some clarifications about previous notes concerning duplicate field names in a result set.



Consider the following relations:



TABLE_A(id, name)

TABLE_B(id, name, id_A)



Where TABLE_B.id_A references TABLE_A.id.



Now, if we join these tables like this: "SELECT * FROM TABLE_A, TABLE_B WHERE TABLE_A.id = TABLE_B.id_A", the result set looks like this: (id, name, id, name, id_A).



The behaviour of mysql_fetch_object on a result like this isn't documented here, but it seems obvious that some data will be lost because of the duplicate field names.



This can be avoided, as Eskil Kvalnes stated, by aliasing the field names. However, it is not necessary to alias all fields on a large table, as the following syntax is legal in MySQL: "SELECT *, TABLE_A.name AS name_a, TABLE_B.name AS name_b FROM TABLE_A, TABLE_B ...". This will produce a result set formatted like this: (id, name, id, name, id_A, name_a, name_b), and your data is saved. Hooray!



-q

rcoles at hotmail dot com
16-Oct-2003 02:11


In reviewing Eskil Kvalnes's comments (04-Mar-2003 11:59 

When using table joins in a query you obviously need to name all the fields to make it work right with mysql_fetch_object()) I was left asking and, as a newbie, the reason why I'm here. I have a 28 field table. Ran SELECT * with a LEFT JOIN, etc and it appears to have worked on my test server without issue.



On further reading, MYSQL.COM has the following:

* It is not allowed to use a column alias in a WHERE clause, because the column value may not yet be determined when the WHERE clause is executed. See section A.5.4 Problems with alias. 

* The FROM table_references clause indicates the tables from which to retrieve rows. If you name more than one table, you are performing a join. For information on join syntax, see section 6.4.1.1 JOIN Syntax. For each table specified, you may optionally specify an alias. 



Aware of the fact there's a difference between tables and fields there appears to be confusion here somewhere.

zhundiak at comcast dot net
15-May-2003 06:44


Here is a wrapper that will allow specifying a class name.

function &db_fetch_object($set,$className)

{

  /* Start by getting the usual array */

  $row = mysql_fetch_assoc($set);

  if ($row === null) return null;



  /* Create the object */

  $obj =& new $className();



  /* Explode the array and set the objects's instance data */

  foreach($row as $key => $value)

  {

    $obj->{$key} = $value;

  }

  return $obj;

}

class CPerson

{

  function getFullName()

  {

    return $this->fname . ' ' . $this->lname;

  }

}

$set = mysql_query('SELECT fname,lname FROM person');

while($person =& db_fetch_object($set,'CPerson'))

{

  echo $person->getFullName();

}

Eskil Kvalnes
03-Mar-2003 10:59


When using table joins in a query you obviously need to name all the fields to make it work right with mysql_fetch_object().

kalleanka
30-Jan-2003 03:50


an addition to the previous... 



for example getting members from a database:



function getAllMembers () {

    $query = "SELECT * FROM people ORDER BY lname";

    $result = mysql_query($query);

            

    while($member = mysql_fetch_object($result)){

        $members[] = $member;

    }

                    

    return $members;

}

<br><br>

    DON'T FORGET TO DECLARE THE ARRAY. If you try to cycle through members after the function has been called and you don't declare the array first you will get a horribly (HORRIBLY!) ugly error in your page. Also, if you try to add the object into the members array inside the while condition instead of in the while loop, you will generate one extra empty space in the array due to the last iteration/check.

spamme at aol dot com
16-Jan-2003 12:28


This is probably a little more elegant:



$sql = "SELECT * FROM table ";

$result = mysql_query($sql);



$data = array();



while ($row = mysql_fetch_object($result))

   $data[] = $row;

allen at brooker dot gb dot net
19-Nov-2002 04:14


I found the above code to be buggy, not adding all the records to the array. This is the code I used instead:



  $command = "SELECT * FROM table ";

  $result = mysql_query($command, $link_id);

  $num = mysql_num_rows($result);



  $clickthru = array();



  for ($i = 0; $i <= $num; $i++) {

    $clickthru[$i] = array();

    $clickthru[$i] = mysql_fetch_array($result);

  }



Allen

Federico at Pomi dot net
15-Sep-2002 07:41


Be carefull:

the object returned will be a new/fresh object.



You can't use this function to replace some attributes of an existing object keeping the old ones. 



Example:

class person

{

   var $name;

   var $surname;

   var $doh;



function print()

{

   print($name." ".$surname);

}



function get_from_db()

{

   $res=query("select name, surname from ppl where... limit 1");

   $this=mysql-fetch-object($res);

}



}



This won't work! When the method get_from_db() is executed, your old object will be destroyed... you won't find anything in the attribute $doh, and if you'll try to call the method print(), it will say it doesn't exist.

amackenz at cs dot uml dot edu
14-Jan-2001 11:03


When selecting with a count/sum, the field must be named.





select count(*) from users;


becomes


select count(*) as total from users;





This way the result can be referenced as:


$row->total;