Question

我在我的代码中添加了这个sql，

function devenir_client_dataforform() {
        $type = $_POST['clientType'];
        //$produitname = $_POST['produitname'];
        $produitnid = $_POST['produitnid'];
        $query = db_select('node', 'n');

        $query->leftjoin('field_data_field_description_col_1', 'img', 'img.entity_id = n.nid');
        $query->leftjoin('field_data_field_pdf_file_image', 'pdf', 'pdf.entity_id = n.nid');
        $query->leftjoin('taxonomy_term_data', 't', 't.tid ='.$type); 



        $query->condition('n.nid', $produitnid, '=')
              //->condition('t.tid', $type, '=')
              ->fields('n', array('title'))
              ->fields('t', array('name'))
              ->fields('pdf', array('field_pdf_file_image_fid'))
              ->fields('img', array('field_description_col_1_value'));

        $result = $query->execute();
        foreach ($result as $record) {
            if(!empty($record->field_description_col_1_value)) {
                $fid =  $record->field_description_col_1_value;
                $produitname =  $record->title;
                $pdf =  $record->field_pdf_file_image_fid;
                $tp = $record->name;

            }
        }   
        $file = file_load($pdf);
        $uri = $file->uri;

        $path = file_create_url($uri);
        //$uri = image_style_path( $path);
         $items[] =  array(
                            "type" => $tp ,
                            "produitname" => $produitname,
                            "produitnid" => $produitnid,
                            "image" => $fid,
                            "pdf" => $path,
                            "typeid" => $type
                    );
        return $items;
    }

但在最后一步出现错误

PDOException：SQLSTATE [42000]：语法错误或访问冲突：1064 您的SQL语法有错误;检查手册对应于您的MySQL服务器版本，以便使用正确的语法在第4行'WHERE（n.nid = NULL）'附近：SELECT n.title AS title，t.name AS名称，pdf.field_pdf_file_image_fid AS field_pdf_file_image_fid， img.field_description_col_1_value AS field_description_col_1_value FROM {node} n LEFT OUTER JOIN {field_data_field_description_col_1} img ON img.entity_id = n.nid LEFT OUTER JOIN {field_data_field_pdf_file_image} pdf ON pdf.entity_id = n.nid LEFT OUTER JOIN {taxonomy_term_data} t ON t.tid = WHERE（n.nid = ：db_condition_placeholder_0）;数组（[：db_condition_placeholder_0] =＆GT; ）在devenir_client_dataforform（）中（/home/dotit_bh/bh-website.dotit.mobi/sites/all/modules/devenir_client/devenir_client.module的第189行）。

Answer 1

SQL Query的这一部分出错：

 WHERE (n.nid = NULL)

更正SQL查询：

WHERE (n.nid is NULL)

你应该重写你的剧本：

$nid_operator = ($produitnid ? '=' : 'is');
$query->condition('n.nid', $produitnid, $nid_operator)
              //->condition('t.tid', $type, '=')
              ->fields('n', array('title'))
              ->fields('t', array('name'))
              ->fields('pdf', array('field_pdf_file_image_fid'))
              ->fields('img', array('field_description_col_1_value'));

语法错误或访问冲突php sql

1 个答案: