Question

我需要你的帮助..

我正在尝试从两个表中检索数据并使用php + mysql插入另一个表，但它不起作用。它向我显示了这条消息（查询有问题）。

这是我的代码：

$emp_id = $_SESSION['emp_id'];

$from= "select department.name from department,employee where emp_id='$emp_id' and department.dept_id = employee.dept_id ";
$result_form = mysql_query($from);

$dept_from = mysql_fetch_assoc($result_form);
$dept_name = $dept_from['department.name'];

$ query =“INSERT INTO Student（date，description，from，emp_id，to）

VALUES

（now（），'$ _ POST [description]'，'$ dept_name'，'$ emp_id'，'$ _ POST [to]'）“;

$result = mysql_query($query);                                          
if(!$result)
    {die("Query got problem").(mysql_error());}
else{

Answer 1

反引号可能会在这里发挥作用。

$query = "INSERT INTO Student (`date`, `description`, `from`, `emp_id`, `to`)
    VALUES
(now(),'$_POST[description]','$dept_name','$emp_id','$_POST[to]')";

Answer 2

试试这个：

你的mysql查询没有正确使用，如下所示：

    $emp_id = $_SESSION['emp_id'];    

    $from= "SELECT d.name FROM department d LEFT JOIN employee e ON d.dept_id = e.dept_id WHERE emp_id = '$emp_id' ";

    $result_form = mysql_query($from);
    $dept_from = mysql_fetch_assoc($result_form);
    $dept_name = $dept_from['name'];

    $query = "INSERT INTO Student (`date`, `description`, `from`, `emp_id`, `to`) VALUES (now(),'".$_POST[description]."','".$dept_name."','".$emp_id."','".$_POST[to]."')"; 

   $result = mysql_query($query);

如果您需要进一步的帮助，请告诉我。

Answer 3

试试这个

 $query = "INSERT INTO Student (date, description, from, emp_id, to)

        VALUES

    (now(),' " . $_POST['description'] ."','$dept_name','$emp_id','". $_POST['to']. "')";

Answer 4

下面的工作代码

$emp_id = 1;

$from= "select department.name from test.department,test.employee where emp_id='$emp_id' and department.dept_id = employee.dept_id ";
$result_form = mysql_query($from);
$dept_from = mysql_fetch_assoc($result_form);

$dept_name      = $dept_from['name'];
$date           = date("Y-m-d H:i:s");
$description    = isset($_POST[description])?$_POST[description]:"none";
$to             = isset($_POST[to])?$_POST[to]:"none";

$query = sprintf("INSERT INTO `test`.`test`
                    (`date`,
                    `description`,
                    `from`,
                    `emp_id`,
                    `to`)
                    VALUES
                    (
                    '%s',
                    '%s',
                    '%s',
                    '%s',
                    '%s'
                    );
                    ",
                    $date,$description,$dept_name,$emp_id,$to); 


$result = mysql_query($query);   

if(!$result)
    {
        die("Query got problem").(mysql_error());
    }

从2个表中获取数据并插入另一个表中

4 个答案: