Question

当我运行我的代码时，我没有错误，也没有发送到数据库，我似乎无法弄清楚问题可能在这里？

我是这个论坛的新手，也是mysql和php的新手，我不确定当你有多个表填写时这是否是插入数据的正确方法

或者它可能与错误的html输入属性有关？

$db = mysql_connect($dbhost, $dbusername, $dbpass);

        $db_select = mysql_select_db($dbdatabase, $db);
          if (!$db_select) {
           die ("Unable to select database: " . mysql_error());
          }

          $query = "SELECT * FROM members, login, skills, indivoffers";
          $result = mysql_query($query);

         if (isset($_POST['mrmrs'],$_POST['fname'],$_POST['lname'],$_POST['gender'],$_POST['addr1'],
                   $_POST['addr2'],$_POST['city'],$_POST['postcode'],$_POST['hometel'],$_POST['mobtel'],
                   $_POST['email'],$_POST['job'],$_POST['user'],$_POST['pass'],$_POST['skill1'],
                   $_POST['skill2'],$_POST['skill3'],$_POST['skill4'],$_POST['skill5'],$_POST['skill6'],
                   $_POST['skill7'],$_POST['skill8'],$_POST['skill9'],$_POST['ortitle'],$_POST['message'],
                   $_POST['offereq'],$_POST['cost'],$_POST['pay'])){
          $title = $_POST['mrmrs'];
          $name = $_POST['fname'];
          $name2 = $_POST['lname'];
          $gender = $_POST['gender'];
          $address1 = $_POST['addr1'];
          $address2 = $_POST['addr2'];
          $city = $_POST['city'];
          $pc = $_POST['postcode'];
          $telhome = $_POST['hometel'];
          $telmob = $_POST['mobtel'];
          $email = $_POST['email'];
          $job = $_POST['job'];

         $username = $_POST['user'];
         $password = $_POST['pass'];

        $skill1 = $_POST['skill1'];
        $skill2 = $_POST['skill2'];
        $skill3 = $_POST['skill3'];
        $skill4 = $_POST['skill4'];
        $skill5 = $_POST['skill5'];
        $skill6 = $_POST['skill6'];
        $skill7 = $_POST['skill7'];
        $skill8 = $_POST['skill8'];
        $skill9 = $_POST['skill9'];

         $titleor = $_POST['ortitle'];
         $mess = $_POST['message'];
         $offerequest  = $_POST['offereq'];
         $cost = $_POST['cost'];
         $pay = $_POST['pay'];

         $sql = "INSERT INTO members (Mr/Mrs, fname, lname, gender, DOB, addr1, addr2, city, postcode, telnohome, telnomob, email, job)
            VALUES ('$title','$name','$name2', '$gender', '$address1', '$address2', '$city', '$pc', '$telhome', '$telmob', '$email', '$job')";
        $letsid = mysql_insert_id( $db);

        $sql = "INSERT INTO login (letsID,username, password)
          VALUES (letsID),'$username','$password')";
        $letsid = mysql_insert_id( $db);

        $sql = "INSERT INTO skills (letsID, skill1, skill2, skill3, skill4, skill5, skill6, skill7, skill8, skill9)
          VALUES (letsID,'$skill1', '$skill2', '$skill3', '$skill4', '$skill5', '$skill6', '$skill7', '$skill8','$skill9')";
        $letsid = mysql_insert_id( $db);

        $sql = "INSERT INTO indivoffers (letsID, title, message, offer/request, cost, pay)
          VALUES (letsID,'$titleor','$mess', '$offerequest', '$cost', '$pay')";
        $letsid = mysql_insert_id( $db);
         }

     ?>

Answer 1

您正在正确创建查询，但实际上并没有执行它们。

所以不要做<?php $sql = "INSERT INTO ..."; ?>而是做这样的事情

<?php $sql = mysql_query("INSERT INTO ..."); ?>

将变量设置为mysql查询将执行查询，因此现在应运行查询。

尝试将多个表单数据插入到数据库的多个表中，没有给出错误但没有工作

1 个答案: