MySQL转义空/空记录

Question

我的数据库中有两个表。一个是 jobs，另一个是 type_of_service，其中绑定(bind)了列 case_reference 和 ref_number，这意味着对于每个案例 case_reference相同的数字出现在 ref_number 中，它是唯一的，并且以自定义方式自动增加，它们看起来像:

JOBS
==========

case_reference | customer | vessel
===============|==========|========
    1311/2     |   Dave   |  rg8   
    1311/3     |   Billy  |  fjg32 
    1311/4     |   Alex   |  sh599

和:

TYPE_OF_SERVICE
===============

ref_number | one | two | three | four 
========================================
   1311/2  |fire |     |       | medical
   1311/3  |     |foam |       |          
   1311/4  |     |foam |engine | medical

现在在我的网络应用程序中，我想以如下方式显示搜索结果:

reference | customer | vessel | type of service
================================================
  1311/2  |  Dave    | rg8    | fire-medical
  1311/3  |  Billy   | fjg32  | foam
  1311/4  |  Alex    | sh599  | foam-engine-medical

我得到结果的方法是:

$记录=数组();

// build array of field names=============================================================================
        $fields=array('customer','vessel',
                     'one','two','three','four');

        // initialize empty array for WHERE clauses
        $wheres=array();

        // loop through field names, get POSTed values,
        // and build array of WHERE clauses, excluding false values
        foreach ($fields as $field) {
          // get existing field value from POST, mark missing or empty value as FALSE
          ${$field} = isset($_POST[$field]) && trim($_POST[$field])!=''
              ? trim($_POST[$field]) : false;

          // add to array of WHERE clauses only if value is not FALSE
          if (${$field}) { $wheres[]="$field LIKE '%".${$field}."%'"; }

        }
        // build SELECT statement from WHERE clauses
        $sql="SELECT * FROM jobs,services WHERE ref_number = case_reference AND ".
             (!empty($wheres) ? implode(" AND ",$wheres) : '1=1').
             ";";

以及显示它们的方法:

<tbody>
            <?php 
                foreach($records as $r) {
                    $watson = $r->case_reference;
                ?>
                    <tr>
                        <td><?php echo  escape($r->case_reference); ?></td>
                        <td><?php echo  escape($r->customer); ?></td>
                        <td><?php echo  escape($r->vessel); ?></td>
                        <td><?php for ($i=0; $i <= $found; $i++) {  echo  escape(<?php echo '<pre>'.$records[$i]->one.'-'.$records[$i]->two.'-'.$records[$i]->three.'-'.
                                        $records[$i]->four'</pre>'} ?>); ?></td>
                    </tr>
                <?php
                }
                ?>
        </tbody>

但我遇到的问题是，当我尝试这样做时，会发生以下结果:

reference | customer | vessel | type of service
================================================
  1311/2  |  Dave    | rg8    | fire---medical-,
  1311/3  |  Billy   | fjg32  | -foam---
  1311/4  |  Alex    | sh599  | -foam-engine-medical-

是否有一种方法/功能可以实现所需的效果，而不显示额外的 "-" 或表 type_of_service 中空单元格的额外空格？

Answer 1

您可以使用CONCAT_WS来简化它

来自MySQL Documentation

mysql> SELECT CONCAT_WS(',','First name','Second name','Last Name');
    -> 'First name,Second name,Last Name'
mysql> SELECT CONCAT_WS(',','First name',NULL,'Last Name');
    -> 'First name,Last Name'

因此 SQL 查询将是

SELECT a.case_reference AS reference, 
       a.customer AS customer, 
       a.vessel AS vessel, 
       CONCAT_WS('-',b.one,b.two,b.three,b.four) AS types_of_service
FROM JOBS a, 
     TYPES_OF_SERVICE b
WHERE a.case_reference = b.ref_number;

如您所见，如果没有值，我假设该字段包含 NULL

ref_number | one | two | three | four 
========================================
1311/2     |fire | NULL| NULL  | medical

如果不是这种情况，则需要一些额外的工作。