Question

我能够解决这个问题但是，当我回应我的client_id时，有很多重复，我试图把array_unique和distinct分开但没有用，我还能做什么呢？我还希望它显示所有地址，因为有些客户端有多个地址。

 $query1= "Select distinct client_id from client_profile where client_status= 'f'";

        $result1= pg_query($conn2, $query1);

    $clientid = array();
    while($row1 = pg_fetch_array($result1)){
        $id=$row1['client_id'];
        $clientid[]=$id;
        $clientid = array_unique($clientid);

    }

    //$clientid=array_unique($clientid);
    $clientid=implode(',',$clientid);


    $query= "select * from vouchers where client_id IN ($clientid)";

    $result = pg_query($conn,$query);



    ?>
    <!DOCTYPE html>
    <html>
        <head>
            <title>Inactive Clients</title>
    <link href="//netdna.bootstrapcdn.com/bootstrap/3.0.3/css/bootstrap.min.css" rel="stylesheet">
    <link href = "http://fonts.googleapis.com/css?family=Roboto:400">
    </head>
        <body>  
    <table class="table table-sm">
      <thead>
        <tr>
          <th>Client id</th>
          <th>File Name</th>
        </tr>
      </thead>
      <?php 

        while($row = pg_fetch_array($result))
        { 


      ?>

      <tbody>

        <tr>
          <td><?php echo $row['client_id']; ?></td>
          <td><?php echo $row['file_name']; ?></td>


        </tr>
      <?php  }?>  </tbody>
    </table>
    </body> 
    </html>

Answer 1

尝试运行以下查询。还要检查列名是否正确。

select client_id,array_agg(address_Field_column) from vouchers where client_id IN (select distinct client_id from client_profile where client_status= 'f') group by client_id;

我不知道你正在使用哪个postgre sql版本。在mysql中，有一个 group_concat 函数。供您参考See this

Answer 2

您可以使用GROUP BY子句。在这种情况下，您的查询可以读取如下内容：

QUERY 1

<?php    
    $query1  = "SELECT client_id FROM `client_profile` AS CP ";
    $query1 .= " WHERE CP.client_status='f' GROUP BY CP.client_id ";

QUERY 2

<?php    
    $query2  = "SELECT * FROM `vouchers` AS VC ";
    $query2 .= " WHERE VC.client_id IN ($clientid) GROUP BY VC.client_id ";

Answer 3

放入查询的开头“SELECT DISTINCT” 你可能需要不止一次。

如何使我的数组输出清晰？

3 个答案: