Question

$sql = "INSERT INTO deelnemer (Account_ID, BehaaldePunten, School_ID, Icon_ID) VALUES
          ('Account_ID', (SELECT ID FROM account WHERE Gebruikersnaam = '$name'),
          0,
          ('School_ID', (SELECT ID FROM school WHERE Naam = '$school'),
          ('Icon_ID', (SELECT ID FROM icon WHERE ID = 1))
          INNER JOIN account ON deelnemer.Account_ID = account.ID,
          INNER JOIN school ON deelnemer.School_ID = school.ID";
          $result = $connection->prepare($sql);
          $result->execute();

嘿，伙计们，我对SQL很陌生。我需要保存一行包含一些外键并且我已经尝试了一些东西，但似乎没有任何效果。我希望你能帮助我xX

这是错误消息：

致命错误：未捕获的异常＆＃39; PDOException＆＃39;与消息＆＃39; SQLSTATE [42000]：语法错误或访问冲突：1064您有 SQL语法错误;查看与您的手册相对应的手册 MariaDB服务器版本，用于在＆JOAH; INNER JOIN附近使用正确的语法帐户ON deelnemer.Account_ID = account.ID，INNER JO＆＃39;在第6行＆＃39;在 C：\ xampp \ htdocs \ GameBattle \ test \ Register_Modal.php：36堆栈跟踪：＃0 C：\ XAMPP \ htdocs中\ GameBattle \测试\ Register_Modal.php（36）： PDOStatement-＆gt; execute（）＃1 C：\ XAMPP \ htdocs中\ GameBattle \测试\ login.php中（84）： require_once（＆＃39; C：\ xampp \ htdocs ...＆＃39;）＃2 {main}引入第36行的C：\ xampp \ htdocs \ GameBattle \ test \ Register_Modal.php

Answer 1

尝试替换：

('Account_ID', (SELECT ID FROM account WHERE Gebruikersnaam = '$name')
          0,
          ('School_ID', (SELECT ID FROM school WHERE Naam = '$school'),

只是：

(SELECT ID FROM account WHERE Gebruikersnaam = '$name',
          0,
 SELECT ID FROM school WHERE Naam = '$school'
etc... )

如何保存外键（SQL）

1 个答案: