ham*_*obi 9 php mysql search zend-framework zend-framework2
我正在创建一个高级搜索,并想通过将它们添加到一个数组来循环我的查询:
private $searchFields = [
'as_first_name' => 'users.first_name like "%VALUE%"',
'as_last_name' => 'users.last_name like "%VALUE%"',
'as_payment_history_invoice_num' => 'users.user_id = (SELECT user_id from payment_history where payment_history.invoice_number = "VALUE" LIMIT 1)',
'as_building_num' => 'property_units.building_number like "%VALUE%"',
'as_residents_email' => 'users.email like "%VALUE%"',
'as_property_name' => 'property.name like "%VALUE%"',
'as_phone_num' => 'REPLACE(REPLACE(REPLACE(REPLACE(users.phone, " ", ""), "(", ""), ")", ""), "-", "") = "VALUE"',
'as_unit_num' => 'property_units.unit_number = "VALUE"',
'as_account_status' => 'user_status.status_name = "VALUE"'
];
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所以在搜索我做的事情...
if (array_key_exists($key, $this->searchFields)) {
$form->get($key)->setValue($val);
$where->NEST->literal(str_replace('VALUE', urldecode($val), $this->searchFields[$key]))->UNNEST;
}
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但问题是我没有逃避任何事情.不好.我怎样才能使用相同的结构,但也可以逃避.
Literal谓词是针对没有占位符的情况.你应该使用Expression谓词.
private $searchFields = [
'as_first_name' => 'users.first_name like "?"',
'as_last_name' => 'users.last_name like "?"',
'as_payment_history_invoice_num' => 'users.user_id = (SELECT user_id from payment_history where payment_history.invoice_number = "?" LIMIT 1)',
'as_building_num' => 'property_units.building_number like "?"',
'as_residents_email' => 'users.email like "?"',
'as_property_name' => 'property.name like "?"',
'as_phone_num' => 'REPLACE(REPLACE(REPLACE(REPLACE(users.phone, " ", ""), "(", ""), ")", ""), "-", "") = "?"',
'as_unit_num' => 'property_units.unit_number = "?"',
'as_account_status' => 'user_status.status_name = "?"'
];
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zend-form值应该已经被解码,因此不需要urldecode
if (array_key_exists($key, $this->searchFields)) {
$form->get($key)->setValue($val);
$where->NEST->expression($this->searchFields[$key], $val)->UNNEST;
}
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我很长一段时间没有使用zend-db,请确保检查此代码实际上是否产生了您需要的查询.