我使用的是zend框架。我在zend中使用了下面的查询,它非常适合我。
$table = $this->getDbTable();
$select = $table->select();
$select->where('name = ?', 'UserName');
$rows = $table->fetchAll($select);
现在我想用'like‘关键字在zend中创建另一个查询。在简单的SQL中是这样的。
SELECT * FROM Users WHERE name LIKE 'U%'
现在如何为上面的查询转换我的zend代码?
发布于 2009-12-17 20:48:39
尝试:
$table = $this->getDbTable();
$select = $table->select();
$select->where('name LIKE ?', 'UserName%');
$rows = $table->fetchAll($select);
或者如果UserName是一个变量:
$table = $this->getDbTable();
$select = $table->select();
$select->where('name LIKE ?', $userName.'%');
$rows = $table->fetchAll($select);
发布于 2013-10-03 17:59:16
$user = new Application_Model_DbTable_User();
// User List
$uname=$_POST['uname'];
$query = $user
->select()
->where('firstname LIKE ?', $uname.'%')
->ORwhere('lastname LIKE ?', $_POST['lname'].'%')
->ORwhere('emailid LIKE ?', $_POST['email'].'%');
$userlist = $user->fetchAll($query);
https://stackoverflow.com/questions/1921394
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