Question

我有以下代码，通过保存表单中输入的内容，从表单发布到数据库。这有效，但我想知道我怎么能够检查我输入的记录是否已经存在，然后将其添加为dublicate并且错误。在这种情况下，应检查student_id是否已存在。如果它存在则应该回显（记录已经存在）

$error1='Add New Intern ';
$error0='No error';

    if(isset($_POST['btnaddint']))
{
    $student_id = trim($_POST['student_id']);
    $comp_name = trim($_POST['comp_name']);
    $comp_supervisor = trim($_POST['comp_supervisor']);
    $comp_tel = trim($_POST['comp_tel']);
    $comp_address = trim($_POST['comp_address']);
    $comp_city = trim($_POST['comp_city']);
    $intake_date = trim($_POST['intake_date']);
    $ass_status = trim($_POST['ass_status']);

    if($student_id == '' || $comp_name == '' || $comp_supervisor == '' || $comp_tel == '' || $comp_address == '' || $comp_city == '' || $intake_date == '' || $ass_status == '')
    {
    $error1=" ERROR - Please make sure all required fields are filled ";

    }
else
    {
    require("server/db.php");
    $tbl_name="int_company"; // Table name 
    // Connect to server and select database.
    mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
    mysql_select_db("$db_name")or die("cannot select DB");

    $student = mysql_query("INSERT INTO $tbl_name (student_id, comp_name, comp_supervisor, comp_tel, comp_address, comp_city, intake_date, ass_status) VALUES('".$student_id."','".$comp_name."','".$comp_supervisor."','".$comp_tel."','".$comp_address."','".$comp_city."','".$intake_date."','".$ass_status."')") or die("Query failed:4 ".mysql_error());
    $error1=" Record has been added... ";
    }
}

Answer 1

试试这个

 <?php
    $error1='Add New Intern ';
    $error0='No error';

    if(isset($_POST['btnaddint']))
    {
        $student_id = trim($_POST['student_id']);
        $comp_name = trim($_POST['comp_name']);
        $comp_supervisor = trim($_POST['comp_supervisor']);
        $comp_tel = trim($_POST['comp_tel']);
        $comp_address = trim($_POST['comp_address']);
        $comp_city = trim($_POST['comp_city']);
        $intake_date = trim($_POST['intake_date']);
        $ass_status = trim($_POST['ass_status']);

        if($student_id == '' || $comp_name == '' || $comp_supervisor == '' || $comp_tel == '' || $comp_address == '' || $comp_city == '' || $intake_date == '' || $ass_status == '')
        {
            $error1=" ERROR - Please make sure all required fields are filled ";

        }
        else
        {
            require("server/db.php");
            $tbl_name="int_company"; // Table name 
            // Connect to server and select database.
            mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
            mysql_select_db("$db_name")or die("cannot select DB");

            $res_student = mysql_query("SELECT student_id FROM $tbl_name WHERE student_id='$student_id' LIMIT 1 ") or die(mysql_error());
            if($row_student = mysql_fetch_assoc($res_student))
            {
                $error1 = "Record is already exists ... ";
            }
            else
            {
                $student = mysql_query("INSERT INTO $tbl_name (student_id, comp_name, comp_supervisor, comp_tel, comp_address, comp_city, intake_date, ass_status) VALUES('".$student_id."','".$comp_name."','".$comp_supervisor."','".$comp_tel."','".$comp_address."','".$comp_city."','".$intake_date."','".$ass_status."')") or die("Query failed:4 ".mysql_error());
                $error1=" Record has been added... ";
            }   
        }
    }

?>

Answer 2

此问题已在stackoverflow中提出，您可以在Logic for already exist record check but only in case of updated form values和using conditional logic : check if record exists; if it does, update it, if not, create it查看此问题。以上链接将帮助您获得答案。我想这会对你有帮助

Answer 3

如果student id字段是UNIQUE索引，则尝试添加相同的id将失败。

然后您可以捕获错误，如果它是由现有记录引起的，则显示您想要的任何内容。

但是如果你正在学习PHP编程，你不应该使用mysql_，因为它已被弃用 - 要么使用mysqli，要么使用PDO

Answer 4

 $student_id =mysql_real_escape_string( trim($_POST['student_id']));
$res = mysql_query('select count(*) from $tbl_name where student_id= ' .$student_id) or die();
$row = mysql_fetch_row($res);
if ($row[0] > 0)
{
    //student_id exists
}
else
{
    //It doesn't
}

在创建新记录之前检查记录是否已存在

4 个答案: