update zend db with multiple where

Question

Is something wrong with my query when i am executing getting fatal error.

$select = $db->select()
    ->from('test', '*')
    ->where('user_id = ?', 1)
    ->where('url is NULL');
$Details = $db->fetchAll($select);

foreach ($Details as $row1) {
    $PublicId = $row1['public_id'];
    $data = array('password' => $randomString , 'flag' => 1);

    $db->update("test", $data, 'public_id =' . $PublicId);
    $pass = $row1['password']; 
}

Error is:

Fatal error: Uncaught exception 'PDOException' with message 'SQLSTATE[42S22]: Column not found: 1054 Unknown column 'wdefabcghbcla45' in 'where clause'' in C:\xampp\php\PEAR\Zend\Db\Statement\Pdo.php:228 Stack trace: #0 C:\xampp\php\PEAR\Zend\Db\Statement\Pdo.php(228): PDOStatement->execute(Array) #1 C:\xampp\php\PEAR\Zend\Db\Statement.php(300): Zend_Db_Statement_Pdo->_execute(Array) #2 C:\xampp\php\PEAR\Zend\Db\Adapter\Abstract.php(479): Zend_Db_Statement->execute(Array) #3 C:\xampp\php\PEAR\Zend\Db\Adapter\Pdo\Abstract.php(238): Zend_Db_Adapter_Abstract->query('UPDATE test ...', Array) #4 C:\xampp\php\PEAR\Zend\Db\Adapter\Abstract.php(634): Zend_Db_Adapter_Pdo_Abstract->query('UPDATE test ...', Array) #5 D:\Zend Workspace\Test\testDelete.php(39): Zend_Db_Adapter_Abstract->update('test', Array, 'test_public_id...') #6 {main} Next exception 'Zend_Db_Statement_Exception' with message 'SQLSTATE[42S22]: Column not found: 1054 Unknown column 'wdefabcghbcla45' in 'where clause'' in C:\xampp\php\PEAR\Zend\Db\Statement\ in C:\xampp\php\PEAR\Zend\Db\Statement\Pdo.php on line 234

Could you try to diplay your arrays? For example, it's not clear value of $PublicId, $randomString? Where error do you get? — Oleg
– Oleg, Commented Nov 29, 2011 at 11:01
i am getting $publicId from query and $randomString from function where i am generating 5 digit random string — vvr
– vvr, Commented Nov 29, 2011 at 11:06
What are their values when you get an error and what type are these fields? — Oleg
– Oleg, Commented Nov 29, 2011 at 11:12
It looks like you try to use genhjllafkvy6u5 column it where which doesn't exists in table. What line causes an error? — Oleg
– Oleg, Commented Nov 29, 2011 at 11:25

Yes Barry · Accepted Answer · 2011-11-30 06:02:57Z

3

The problem is your WHERE clause was ending up like WHERE public_id = wdefabcghbcla45 therefore MySQL was trying "wdefabcghbcla45" as a column instead of a value. You need to wrap it with quotes (single or double..doesn't matter).

Try it like this:

$where = $db->quoteInto('public_id = ?', $PublicId);
$db->update('test', $data, $where);

http://files.zend.com/help/Zend-Framework/zend.db.html#zend.db.adapter.quoting.quote-into

answered Nov 30, 2011 at 6:02

Yes Barry

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Comments

mymotherland · Accepted Answer · 2011-11-29 11:12:43Z

1

Try like

$data = array('password' => $randomString , 'flag' => 1);
$db->update( "test", $data, 'public_id ='.$PublicId );

answered Nov 29, 2011 at 11:12

mymotherland

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Comments

gprathour · Accepted Answer · 2011-11-30 05:38:28Z

0

When I tried like this, it worked. Thanks for helping me.

$data = array('password' => $randomString , 'flag' => 1);
$where = "public_id = '" . $PublicId."'";
$db->update( "test", $data, $where );

edited Nov 30, 2011 at 5:38

gprathour

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answered Nov 30, 2011 at 4:24

vvr

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