Question

我是mysql的新手，我正在尝试创建一个可以在一个表上存储用户电子邮件和密码的数据库以及他们在另一个表上输入的值，如何加入表以确保输入的值链接到正确的用户。这是我一直在使用的代码，但它不允许在运行外键时存储该值，但如果我删除外键，我可以存储该值。请帮忙。

CREATE TABLE IF NOT EXISTS `data` (
  `user_id` int(11) NOT NULL AUTO_INCREMENT,
  `email` varchar(51) NOT NULL,
  `password` varchar(15) NOT NULL,
  PRIMARY KEY (`user_id`),
  UNIQUE KEY `email_UNIQUE` (`email`)
)


CREATE TABLE IF NOT EXISTS `gluco` (
  `G_id` int(11) NOT NULL AUTO_INCREMENT,
  `bloods` decimal(4,2) NOT NULL,
  `time` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
  `user_id` int(11) NOT NULL,
    FOREIGN KEY (`user_id`) REFERENCES `data`(`use_id`),
    UNIQUE KEY `G_id_UNIQUE` (`G_id`)
)


<?php
include('db.php'); 

if (!isset($_POST['reading'])) { //checking if user has entered this page directly
    include('contactus.php');
} else {
    if (isset($_POST['reading'])&&$_POST['reading']==""||!isset($_POST['reading'])) {
        $error[] = "fill in your blood/glucose"; 
    }

    $reading = mysql_real_escape_string($_POST['reading']);
    $sql = "SELECT * FROM gluco WHERE bloods  = '$reading'";

    if(isset($error)){
        if(is_array($error)){
            echo "<div class=\"error\"><span>please check the errors and refill the form<span><br/>";

            foreach ($error as $ers) {
                echo "<span>".$ers."</span><br/>";
            }

            echo "</div>";

            include('contactus.php');
        }
    }

    if(!isset($error)){
        $sreading=mysql_real_escape_string($_POST['reading']);
        $sip=mysql_real_escape_string($_SERVER['HTTP_HOST']);
        $save = mysql_query("INSERT INTO `gluco` ( `bloods`  )VALUES ('$sreading')");

        if($save){
            echo "<div class=\"success\"><span>Your reading has been successfully stored</span><br/></div>";
        } else {
            echo "<div class=\"warning\"><span>Some Error occured during processing your data</div>";
        }
    }
}

?>

Answer 1

你的代码逻辑正确。但是引用的列名称出现错误：

CREATE TABLE IF NOT EXISTS `data` (
  `user_id` int(11) NOT NULL AUTO_INCREMENT,
  `email` varchar(51) NOT NULL,
  `password` varchar(15) NOT NULL,
  PRIMARY KEY (`user_id`),
  UNIQUE KEY `email_UNIQUE` (`email`)
)


CREATE TABLE IF NOT EXISTS `gluco` (
  `G_id` int(11) NOT NULL AUTO_INCREMENT,
  `bloods` decimal(4,2) NOT NULL,
  `time` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
  `user_id` int(11) NOT NULL,
  FOREIGN KEY (`user_id`) REFERENCES `data`(`user_id`),
  UNIQUE KEY `G_id_UNIQUE` (`G_id`)
)

并在这一行：

$save = mysql_query("INSERT INTO `gluco` ( `bloods`  )VALUES ('$sreading')");

您没有在insert语句中设置user_id，因此，外键将不起作用，并且不会进行插入。因此，您需要将用户ID存储在变量中（因为我不知道代码中的上下文和范围，我无法帮助您设置此变量）。所以，你的代码应该是这样的：

$save = mysql_query("INSERT INTO gluco (bloods, user_id)VALUES ('$sreading', $user_id)");

试图在mysql中的不同表中连接某些值

1 个答案: