php - 如何在PHP中捕获错误代码

Question

我想使用数据库员工的属性ID和empName将记录添加到表'pool'中

theform.html

<FORM NAME ='form2' METHOD ='POST' ACTION ='result.php' enctype="multipart/form-data">
<B>Board Write</B> <BR>

<INPUT TYPE = Text Name = empID value = 'write ID here'>
<INPUT TYPE = Text Name = empName  VALUE = 'write name here'><P>
<INPUT TYPE = 'Submit' Name = Submit2  VALUE = 'Post'>

</FORM>

result.php

<?PHP

      $ID = $_POST['empID'];
      $NAME = "'" . $_POST['empName'] . "'";

       $server = "127.0.0.1"; //connect to server
       $user_name = "root";   
       $password = "";

       $database = "employees";


    $db_handle = mysql_connect($server, $user_name, $password);
    $db_found = mysql_select_db($database, $db_handle);


if($db_found){// there is a database

$tableSQL = "INSERT INTO pool(ID, empName) VALUES " . "(" . $ID . "," . $NAME . ")";


$result = mysql_query($tableSQL);

    if($result){
            print "success!";
    }
    else{
            print "fail!";
        }


}

   mysql_close($db_handle);  //close the connection;



?>

在数据库中，ID是唯一的，也是INT数据类型，我想在用户输入数据库中已经存在的ID值的地方捕获错误代码。我们如何做到这一点？

Answer 1

您可以使用 mysql_errno() (如果使用mysqli_errno()，则使用 mysqli )在查询失败后获取错误号(检查 mysql_query() 的返回值)。根据mysql error documentation应该是#1022(ER_DUP_KEY)错误