Question

我查看并测试了几种解决方案，但仍然存在同样的问题..

我有两个表：tb_cadastro（登录）和tb_quest（问卷答案）

    CREATE TABLE IF NOT EXISTS `tb_cadastro` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `id_quest` int(11) NOT NULL,
  `name` varchar(20) NOT NULL,
  `surname` varchar(20) NOT NULL,
  `email` varchar(45) NOT NULL,
  `password` varchar(35) NOT NULL,
  `sex` varchar(4) NOT NULL,
  PRIMARY KEY (`id`),
  KEY `fk_id_quest` (`id_quest`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8 AUTO_INCREMENT=2 ;     
ALTER TABLE `tb_cadastro`
  ADD CONSTRAINT `fk_id_quest` FOREIGN KEY (`id_quest`) REFERENCES `tb_quest` (`id_quest`) ON UPDATE NO ACTION;

 CREATE TABLE IF NOT EXISTS `tb_quest` (
  `id_quest` int(11) NOT NULL AUTO_INCREMENT,
  `q1` varchar(1) NOT NULL,
  `q2` varchar(1) NOT NULL,
  `q3` varchar(1) NOT NULL,
  `q4` varchar(1) NOT NULL,
  `q5` varchar(1) NOT NULL,
  `q6` varchar(1) NOT NULL,
  `q7` varchar(1) NOT NULL,
  `q8` varchar(1) NOT NULL,

  PRIMARY KEY (`id_quest`),
  UNIQUE KEY `id_quest` (`id_quest`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8 AUTO_INCREMENT=4 ;

这两个表与id_quest

相关

问题是tb_cadastro表没有向db添加任何内容，但也没有显示错误.. table tb_quest正常添加数据我也可以通过mysql

在tb_cadastro中添加数据

我得出的结论是问题出现在FK约束中，所以我删除了FK的关系并且一切正常

我尝试了所有我发现的东西，所以如果你有解决方案，我将不胜感激

这是我的PHP代码

<?php
    session_start();

    require_once ("conexao.php");
    mysql_connect("$hostname", "$username", "$senhabd") or die(mysql_error());
    mysql_select_db($banco) or die(mysql_error());


    $_SESSION['Usuario'] = $_POST['Usuario'];
    $_SESSION['S_usuario'] = $_POST['S_usuario'];
    $_SESSION['Email'] = $_POST['Email'];
    $_SESSION['C_email'] = $_POST['C_email'];
    $_SESSION['Senha'] = $_POST['Senha'];
    $_SESSION['C_senha'] = $_POST['C_senha'];
    $_SESSION['Sexo'] = $_POST['Sexo'];

    $_SESSION['q1'] = $_POST['q1'];
    $_SESSION['q2'] = $_POST['q2'];
    $_SESSION['q3'] = $_POST['q3'];
    $_SESSION['q4'] = $_POST['q4'];
    $_SESSION['q5'] = $_POST['q5'];
    $_SESSION['q6'] = $_POST['q6'];
    $_SESSION['q7'] = $_POST['q7'];
    $_SESSION['q8'] = $_POST['q8'];

    $insert="INSERT INTO tb_quest (q1,q2,q3,q4,q5,q6,q7,q8) VALUES( '" . $_SESSION['q1'] . "','" . $_SESSION['q2'] . "','" . $_SESSION['q3'] . "','" . $_SESSION['q4'] . "','" . $_SESSION['q5'] . "','" . $_SESSION['q6'] . "','" . $_SESSION['q7'] . "','" . $_SESSION['q8'] . "')"or die (mysql_error());
    mysql_query($insert);

    $id_quest = mysql_insert_id();
    //echo $id_quest;

    $insert="INSERT INTO tb_cadastro (id_quest,name,surname,email,password,sex) VALUES( $id_quest,'" . $_SESSION['Usuario'] . "','" . $_SESSION['S_usuario'] . "','" . $_SESSION['Email'] . "','" . $_SESSION['Senha'] . "','" . $_SESSION['Sexo'] . "','" . $_SESSION['Sexo'] . "')" or die (mysql_error());
    mysql_query($insert);

?>

Answer 1

我不做PHP或mysql，但我猜你想在SQL字符串之外引用变量$ id_quest。

$insert="INSERT INTO tb_cadastro (id_quest,name,surname,email,password,sex) VALUES(" . $id_quest . ",'" . $_SESSION['Usuario'] . "','" . $_SESSION['S_usuario'] . "','" . $_SESSION['Email'] . "','" . $_SESSION['Senha'] . "','" . $_SESSION['Sexo'] . "','" . $_SESSION['Sexo'] . "')" or die (mysql_error());

使用外键在表中添加数据

1 个答案: