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drupal db_select に問題があります。

これが私のコードです:

    $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();

DBで私のリクエストは

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)

このリクエストは、phpmyadminで試してみると機能しますが、代わりにmysqlログで

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')

最後に、WHERE で、「fpffsfc.feature_support_compared_target_id」が一重引用符で囲まれていることを確認してください。

これは明らかに、->condition の 2 番目の引数が変数のみを受け入れるように見えるためです。db_select を使用して 2 つの db フィールドで条件を作成する方法を知っている人はいますか?

助けてくれてありがとう。

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1 に答える 1

2

$query->where($snippet, $args = array()); を使用します。

$query->where('cfsfc.feature_support_compared_target_id = fpffsfc.feature_support_compared_target_id');
于 2012-10-17T23:06:53.773 に答える