Drupal db_select(),如何在条件中使用两个db字段？

2019年8月4日 194次阅读

我有一个drupal db_select的问题.

这是我的代码：

    $query = db_select('node', 'n');
    $query->addField('n', 'nid', 'nid');
    $query->addField('cfs', 'entity_id', 'feature_support_id');
    $query->addField('fpffs', 'entity_id', 'parent_feature_support_id');
    $query->addField('cfsfc', 'feature_support_compared_target_id', 'feature_support_compared');
    $query->addField('fpffsfc', 'feature_support_compared_target_id', 'parent_feature_support_compared');
    //Get feature_support of the feature
    $query->join('field_data_feature_support_feature', 'cfs', 'n.nid = cfs.feature_support_feature_target_id');
    $query->join('field_data_feature_support_compared', 'cfsfc', 'cfs.entity_id = cfsfc.entity_id');
    //Get parent feature_support through feature
    $query->join('field_data_feature_parent_feature', 'fp', 'n.nid = fp.entity_id');
    $query->join('field_data_feature_support_feature', 'fpffs', 'fp.feature_parent_feature_target_id = fpffs.feature_support_feature_target_id');
    $query->join('field_data_feature_support_compared', 'fpffsfc', 'fpffs.entity_id = fpffsfc.entity_id');
    $query->condition('n.nid', $node_revision->nid, '=');
    $query->condition('cfsfc.feature_support_compared_target_id', 'fpffsfc.feature_support_compared_target_id', '=');
    $result = $query->execute();

在DB我的请求应该是

SELECT n.nid AS nid, cfs.entity_id AS feature_support_id, fpffs.entity_id AS parent_feature_support_id, cfsfc.feature_support_compared_target_id AS feature_support_compared, fpffsfc.feature_support_compared_target_id AS parent_feature_support_compared
FROM node n
INNER JOIN field_data_feature_support_feature cfs ON n.nid = cfs.feature_support_feature_target_id
INNER JOIN field_data_feature_support_compared cfsfc ON cfs.entity_id = cfsfc.entity_id
INNER JOIN field_data_feature_parent_feature fp ON n.nid = fp.entity_id
INNER JOIN field_data_feature_support_feature fpffs ON fp.feature_parent_feature_target_id = fpffs.feature_support_feature_target_id
INNER JOIN field_data_feature_support_compared fpffsfc ON fpffs.entity_id = fpffsfc.entity_id
WHERE  (n.nid = '9') AND (cfsfc.feature_support_compared_target_id = fpffsfc.feature_support_compared_target_id)

当我在phpmyadmin中尝试它时,此请求工作,但我在mysql日志中

SELECT n.nid AS nid, cfs.entity_id AS feature_support_id, fpffs.entity_id AS parent_feature_support_id, cfsfc.feature_support_compared_target_id AS feature_support_compared, fpffsfc.feature_support_compared_target_id AS parent_feature_support_compared
FROM node n
INNER JOIN field_data_feature_support_feature cfs ON n.nid = cfs.feature_support_feature_target_id
INNER JOIN field_data_feature_support_compared cfsfc ON cfs.entity_id = cfsfc.entity_id
INNER JOIN field_data_feature_parent_feature fp ON n.nid = fp.entity_id
INNER JOIN field_data_feature_support_feature fpffs ON fp.feature_parent_feature_target_id = fpffs.feature_support_feature_target_id
INNER JOIN field_data_feature_support_compared fpffsfc ON fpffs.entity_id = fpffsfc.entity_id
WHERE  (n.nid = '9') AND (cfsfc.feature_support_compared_target_id = 'fpffsfc.feature_support_compared_target_id')

最后看,在WHERE中,’fpffsfc.feature_support_compared_target_id’周围有单引号,不应该存在.

这显然是因为 – >条件的第二个参数似乎只接受变量.任何人都知道如何使用db_select创建两个db字段的条件？

感谢您提供的任何帮助.

最佳答案 使用$query-> where($snippet,$args = array());

$query->where('cfsfc.feature_support_compared_target_id = fpffsfc.feature_support_compared_target_id');
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