Nav*_*eed 17 sql zend-framework
我正在使用zend框架.我在zend中使用以下查询,它完全适合我.
$table = $this->getDbTable();
$select = $table->select();
$select->where('name = ?', 'UserName');
$rows = $table->fetchAll($select);
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现在我想在zend中使用'like'关键字创建另一个查询.在简单的SQL中就是这样.
SELECT * FROM Users WHERE name LIKE 'U%'
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现在如何转换我的zend代码以进行上述查询?
Jus*_*sel 43
尝试:
$table = $this->getDbTable();
$select = $table->select();
$select->where('name LIKE ?', 'UserName%');
$rows = $table->fetchAll($select);
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或者如果UserName是变量:
$table = $this->getDbTable();
$select = $table->select();
$select->where('name LIKE ?', $userName.'%');
$rows = $table->fetchAll($select);
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