在MultiSelect属性上使用Find_in_set收集，但有些只有一个值

Question

我正在构建产品集合，该产品将从多选择属性中找到结果。

（将Finset添加到集合中的相关代码 - 调整为显示真实ID值）

$attribute = Mage::getModel('eav/config')->getAttribute('catalog_product', 'measurement');
    $value = array('finset' => array('237',
                '236',
                '235',
                '234',
                '233',));
    $collection->addAttributeToFilter($attribute, $value);

所得的SQL（添加了Visiblity过滤器）就是这样：

SELECT DISTINCT
    e . *,
    at_measurement.value AS measurement,
    at_visibility.value AS visibility
FROM
    catalog_product_entity AS e
        INNER JOIN
    catalog_product_entity_varchar AS at_measurement ON (at_measurement.entity_id = e.entity_id) AND (at_measurement.attribute_id = '983') AND (at_measurement.store_id = 0)
        INNER JOIN
    catalog_product_entity_int AS at_visibility ON (at_visibility.entity_id = e.entity_id) AND (at_visibility.attribute_id = '526') AND (at_visibility.store_id = 0)
WHERE
    (e.attribute_set_id IN ('74')) AND (FIND_IN_SET('237',
            '236',
            '235',
            '234',
            '233',
            at_measurement.value)) AND (at_visibility.value IN ('2' , '4'))
GROUP BY e.entity_id

问题是我得到了一个SQL错误：

"SQLSTATE[42000]: Syntax error or access violation: 1582 Incorrect parameter count in the call to native function 'FIND_IN_SET'"

我（认为）我可以理解原因：某些多选择值只有一个选项，因此没有逗号分隔的值可以符合限定为finset

我为什么会产生此错误？如何编写此集合对象来解释这一点？

如果不是以上，我想念什么？

在MySQL Workbench中运行SQL的结果，减去Find_in_set子句：

Answer 1

尝试使用或条件使用AddAttributetofterter

$collection->addAttributeToFilter($attribute,
    array(
        array('finset'=> array('237')),
        array('finset'=> array('238')),
        array('finset'=> array('239')),
    )
);

或者

$collection->addAttributeToFilter(
    array(
        array('attribute'=> 'attributecode','finset' => array('237')),
        array('attribute'=> 'attributecode','finset' => array('237')),
        array('attribute'=> 'attributecode','finset' => array('237')),
    )
);