Question

我有两个表表联系人和表公司。在表格联系中有4列称为公司ID（1-4）在这些列中，存在与该人员相关联的公司ID。现在的问题是，我希望能够显示公司名称而不是id，我尝试使用以下查询，但它只给出了我最后一行的值

select *
from contact as ct
 join company as cy1 on cy1.Company_id = ct.Company_id1
 join company as cy2 on cy2.Company_id = ct.Company_id2
 join company as cy3 on cy3.Company_id = ct.Company_id2
 join company as cy4 on cy4.Company_id = ct.Company_id4


$stmt = sqlsrv_query( $conn, $sql , array(), array( "Scrollable" => 'static' ) );
if( $stmt === false) {
    die( print_r( sqlsrv_errors(), true) );
}
$stmtcount = sqlsrv_num_rows($stmt);
if ($stmtcount > 0) {
while( $row = sqlsrv_fetch_array( $stmt, SQLSRV_FETCH_ASSOC) ) {
                                    $id = $row['Contact_id'];
                                    $fname = $row['Firstname'];
                                    $lname = $row['Lastname'];                                       
                                    $company1=$row['Company'];
                                    $company2=$row['Company'];
                                    $company3=$row['Company'];
                                    $company4=$row['Company'];



echo '<tr>';
echo "<td> <a class='notice' href='details.php?id=" . $id . " '>" .$row['Abbreviation'].'&nbsp'. $row['Firstname'] . "</a> </td>";

echo '<td>'.$lname.'</td>';
echo '<td>'.$company1.'</td>';
echo '<td>'.$company2.'</td>';
echo '<td>'.$company3.'</td>';
echo '<td>'.$company4.'</td>';

echo '</tr>';

结果如下result of the query 表结构

and here's the database table

Answer 1

尝试使用列别名而不是select *：

select ct.*,
       cy1.name as name1,
       cy2.name as name2,
       cy3.name as name3,
       cy4.name as name4
from contact ct join
     company cy1
     on cy1.Company_id = ct.Company_id1 join
     company
     cy2
     on cy2.Company_id = ct.Company_id2 join
     company cy3
     on cy3.Company_id = ct.Company_id2 join
     company cy4
     on cy4.Company_id = ct.Company_id4;

如果多个列具有相同的名称，则查询将任意选择一个值。

将所有id转换为表格，将其转换为另一个表格中的值

1 个答案: