Question

如何从codeigniter中的连接表中读取多个值

这是model.php

<table class="table table-bordered table-hover" id="dataTables-example">
    <thead>
        <tr>
            <th class="col-sm-1" >VILLAGE</th>
            <th class="col-sm-1" >ZONE</th>
            <th class="col-sm-1" >SCHOOL</th>                    
        </tr>
    </thead>
    <tbody>                    
        <?php if (!empty($all_village_info)):
            foreach ($all_village_info as $v_village) :
                ?>
                <tr>
                    <td><?php echo $v_village->village_name ?></td> 
                    <td><?php echo $v_village->zone_name ?></td>
                    <td><?php echo $v_village->school_name ?></td>                                                                     
                </tr>
                <?php
            endforeach;
            ?>
<?php endif; ?>
    </tbody>
</table>

这是view.php

|-------------|------------|
|village_name |school_id   |
|-------------|------------|
|Bhiwani      |1,2,3       |

sql_is中的

值

    |-------------|------------|
    |Village      |School      |
    |-------------|------------|
    |Bhiwani      |School 1    |

视图为

|-------------|-------------------------------|
|Village      |School                         |
|-------------|-------------------------------|
|Bhiwani      |School 1, School 2, School 3   |

我想表现出像

SecurityModule.forRoot(new SecurityMockService())

我该怎么做？帮助PLZ 如何从codeigniter中的连接表中读取多个值

Answer 1

您可以使用表格的别名，然后选择那些

注意：我强烈建议您使用*当且仅当您使用其他所有列时，它才会使您的查询位置变慢。

/ *你可以为表取一个别名并从中选择*，我遵循链接方法。 * /

Number 1: <input type="text" placeholder="Number 1" id="value1" name="value1" /><br>
Number 2: <input type="text" placeholder="Number 2" id="value2" name="value2" />
<select id="user_select">
    <option value="+" selected="selected">Addition</option>
    <option value="/">Division</option>
    <option value="*">Multiplication</option>
</select> <input type="button" name="Sumbit" value="Calculate" onclick="calculate()" /><br>
Result: <input type="text" id="answer" name="answer" value="Your Answer" />

Answer 2

您可以针对您的问题尝试此解决方案：

请更改您的模态功能：

all_village_info:

public function all_village_info($id = NULL) {
    $this->db->select('tbl_village.*', FALSE);
    $this->db->select('GROUP_CONCAT(tbl_school.school_name ORDER BY tbl_school.school_id) as school_name', FALSE);
    $this->db->from('tbl_village');
    $this->db->join('tbl_school', 'FIND_IN_SET(tbl_school.school_id, tbl_village.school_id) > 0', 'left');
    if (!empty($id)) {
        $this->db->where('tbl_village.village_id', $id);
        $query_result = $this->db->get();
        $result = $query_result->row();
    } else {
        $query_result = $this->db->get();
        $result = $query_result->result();
    }
    return $result;
}

我希望它会对你有所帮助。

如何从codeigniter中的连接表中读取多个值

2 个答案: