Question

我的php / mysql UPDATE查询无法到达我的数据库时遇到问题。我知道这可能有一个简单的修复，我似乎无法找到我的错误，这是我的代码：

这是我用来发送数据的表单：

<?php

        if(!($stmt = $mysqli->prepare("SELECT bowl_games.id, bowl_games.name, stadiums.name, bowl_games.inaugural_year FROM bowl_games
                        INNER JOIN stadiums ON stadiums.id = bowl_games.stadium_id 
                        WHERE bowl_games.id = ? "))){
            echo "Prepare failed: "  . $stmt->errno . " " . $stmt->error;
        }

        if(!($stmt->bind_param("i", $_POST['bowl_game']))){
            echo "Bind failed: "  . $stmt->errno . " " . $stmt->error;
        }

        if(!$stmt->execute()){
            echo "Execute failed: "  . $mysqli->connect_errno . " " . $mysqli->connect_error;
        }
        if(!$stmt->bind_result($id, $name, $stadium, $inauguralyear)){
            echo "Bind failed: "  . $mysqli->connect_errno . " " . $mysqli->connect_error;
        }
        while($stmt->fetch()){

        }
        ?>

            <div class="container">
            <form method="post" action="update_bowl_game_2.php">
            <fieldset> <legend>Update Bowl Game</legend>

                <div class="form-group row">
                <label class="col-sm-2 col-form-label">Name</label>
                <div class="col-sm-10">
                    <input type="text", class="form-control", name="Name", value="<?php echo $name?>"/>
                </div>
                </div>

                <div class="form-group row">
                <label class="col-sm-2 col-form-label">Stadium</label>
                <div class="col-sm-10">
                    <select name="Stadium">
                        <?php 
                        if(!($stmt = $mysqli->prepare("SELECT id, name FROM stadiums ORDER BY name"))){
                            echo "Prepare failed: "  . $stmt->errno . " " . $stmt->error;
                            }

                        if(!$stmt->execute()){
                            echo "Execute failed: "  . $mysqli->connect_errno . " " . $mysqli->connect_error;
                        }
                        if(!$stmt->bind_result($id, $sname)){
                            echo "Bind failed: "  . $mysqli->connect_errno . " " . $mysqli->connect_error;
                        }
                        while($stmt->fetch()){
                            if($sname === $stadium){
                                echo "<option value=\"" .   $id . "\" selected>" . $sname . "</option>";
                                } else {
                                echo "<option value=\"" .   $id . "\">" . $sname . "</option>";
                                }
                        }
                        $stmt->close();
                        ?>
                    </select>
                </div>
                </div>

                <div class="form-group row">
                <label class="col-sm-2 col-form-label">Inaugural Year</label>
                <div class="col-sm-10">
                    <input type="number", class="form-control", name="InauguralYear", value="<?php echo $inauguralyear?>"/>
                </div>
                </div>

                    <input type="hidden" name="id" value="<?php echo $id?>"/>
                <div class="form-group row">
                <div class="offset-sm-2 col-sm-10">
                    <button type="submit" class="btn btn-primary">Update Bowl Game</button>
                </div>
                </div>
            </fieldset>
            </form>
        </div>
<?php
        $mysqli = "SELECT bowl_games.id, bowl_games.name, stadiums.name, bowl_games.inaugural_year FROM bowl_games
                   INNER JOIN stadiums ON stadiums.id = bowl_games.stadium_id"
        ?>

这是应该更新数据库中条目的PHP代码：

<?php
//Turn on error reporting
ini_set('display_errors', 'On');
//Connects to the database
$mysqli = new mysqli("oniddb.cws.oregonstate.edu","dejarnen-db","*hidden*","dejarnen-db");
if(!$mysqli || $mysqli->connect_errno){
    echo "Connection error " . $mysqli->connect_errno . " " . $mysqli-   >connect_error;
    }

if(!($stmt = $mysqli->prepare("UPDATE bowl_games SET name=?, stadium_id=?, inaugural_year=? WHERE id= ?"))){
    echo "Prepare failed: " . $stmt->errno . " " . $stmt->error;
}

if(!($stmt- >bind_param("siii",$_POST['Name'],$_POST['Stadium'],$_POST['InauguralYear'],$_POST['id']))){
    echo "Bind failed: " . $stmt->errno . " " . $stmt->error;
}

if(!$stmt->execute()){
    echo "Execute failed: " . $stmt->errno . " " . $stmt->error;
} else {
    echo "Updated " . $stmt->affected_rows . " rows in bowl games.";
}
?>

当我提交表单时，如果所选条目已成功更新，我应该看到消息＆＃34;在碗游戏中更新了1行。＆＃34;相反，我得到了消息＆＃34;在碗游戏中更新了0行。＆＃34;

任何人都可以指出我正朝这个问题迈进的方向吗？谢谢

Answer 1

在表单中，您将名为"use strict"; var fs = require("fs"); var path = require("path"); var Sequelize = require('sequelize') , sequelize = new Sequelize(process.env.MYSQL_DB, process.env.MYSQL_USER, process.env.MYSQL_PASSWORD, { dialect: "mysql", // or 'sqlite', 'postgres', 'mariadb' port: 3306, // or 5432 (for postgres) }); var db = {}; fs .readdirSync(__dirname) .filter(function(file) { return (file.indexOf(".") !== 0) && (file !== "index.js"); }) .forEach(function(file) { var model = sequelize.import(path.join(__dirname, file)); db[model.name] = model; }); Object.keys(db).forEach(function(modelName) { if ("associate" in db[modelName]) { db[modelName].associate(db); } }); db.sequelize = sequelize; db.Sequelize = Sequelize; module.exports = db;的变量用于两个不同的目的：

对于游戏的身份
对于体育场馆的身份

首先，您将游戏的ID检索到变量中，然后将体育场的ID检索到变量中，然后使用该变量为游戏ID创建隐藏的输入。

当你编写隐藏的输入时，$ id变量包含一个体育场的id。

一种可能的解决方案：列出体育馆时，请使用单独的变量名称：

$id

通常，使用特定的变量名称是个好主意。因此，使用＆＃34; $ stadium_name＆＃34;而不是＆＃34; $ name＆＃34;等等。该规则的唯一例外是当函数中的局部变量非常短时。

另一种可能的解决方案是在填充体育场馆的选择之前，先写下隐藏的输入。

Answer 2

首先检查$_POST数据不是null，然后使用绑定参数sssi。

//example
if(!($stmt- >bind_param("sssi", $_POST['Name'], $_POST['Stadium'], $_POST['InauguralYear'], $_POST['id']))){
echo "Bind failed: " . $stmt->errno . " " . $stmt->error;

PHP更新查询未到达数据库

2 个答案: