Question

我有两张这样的表：

员工

EmployeeID  EmployeeName
1234        Jessica
1235        Tiffany
1236        Kayla
1237        Jackson
1238        Junior
1239        Ray
1240        Raymond

和...

表格

IDForm Form_EmployeeID Content AgreementBy  Verificationby  Validateby  receiver

1      1234            abcde   1235         1236            1237        1240

2      1238            dcbe    1235         1239            1237        1240

问题是，我想显示如下数据：

我尝试过使用此代码：

$id=$_GET['id'];

$sql=mysql_query("SELECT Employee.*, Form.*
                        FROM Form
                        INNER JOIN Employee ON Form.Form_EmployeeID= Employee.EmployeeID
                        where Form.FormID = '$id'") or die(mysql_error()); $data=mysql_fetch_array($sql);

但数据显示如下：

Form ID        : 1
Employee Name  : Jessica
Content        : abcde
Agreement By   : 
Verification By: 
Validate By    : 
Received By    :

或者像这样：

 Form ID        : 1
Employee Name  : Jessica
Content        : abcde
Agreement By   : 1235
Verification By: 1236
Validate By    : 1237
Received By    : 1240

我可以知道错误在哪里吗？

Answer 1

你的查询是这样的......

SELECT Form.*, 
    a.EmployeeName as employee_name,
    b.EmployeeName as agreement_by,
    c.EmployeeName as verification_by,      
    d.EmployeeName as validate_by,
    e.EmployeeName as receiver_by
                    FROM Form
                    LEFT JOIN Employee a ON Form.Form_EmployeeID= a.EmployeeID
                    LEFT JOIN Employee b ON Form.AgreementBy= b.EmployeeID
                    LEFT JOIN Employee c ON Form.Verificationby= c.EmployeeID
                    LEFT JOIN Employee d ON Form.Validateby= d.EmployeeID
                    LEFT JOIN Employee e ON Form.receiver= e.EmployeeID
                    where Form.FormID = '$id'

Answer 2

尝试此查询，不要在选择查询中使用＆＃39; *＆＃39; ...

 SELECT f.IDForm , f.Content ,
 emp1.EmployeeName as employee_name,
 emp2.EmployeeName as agreement_by,
 emp3.EmployeeName as verification_by ,      
 emp4.EmployeeName as validate_by,
 emp5.EmployeeName as receiver_by
 FROM Form f
    INNER JOIN Employee emp1 ON emp1.Form_EmployeeID= f.EmployeeID
    INNER JOIN Employee emp2 ON emp2.AgreementBy= f.EmployeeID
    INNER JOIN Employee emp3 ON emp3.Verificationby= f.EmployeeID
    INNER JOIN Employee emp4 ON emp4.Validateby= f.EmployeeID
    INNER JOIN Employee emp5 ON emp5.receiver= f.EmployeeID
 where Form.FormID = '$id'

如何使用PHP在MySQL表上使用内连接？

2 个答案: