sql-insert 相关问题

我正在尝试使用 PHP 将数据添加到 MySql 数据库中。代码：我正在尝试使用 PHP 将数据添加到 MySql 数据库中。代码： <?php session_start(); include('config.php'); if(isset($_POST['Submit'])) { $city = $_POST['city']; $from_station = $_POST['from_station']; $from_date = $_POST['from_date']; // Prepare and execute the SQL query $sql = "INSERT INTO journeys (city, from_station, from_date) VALUES ('$city', '$from_station', '$from_date')"; $result = $conn->query($sql); if (mysqli_query($conn, $sql)) { echo "New record created successfully"; } else { echo "Error: " . $sql . "<br>" . mysqli_error($conn); } } ?> <!DOCTYPE html> <html lang="pt_PT"> <head> <meta charset="UTF-8"> <meta name="viewport" content="width=device-width, initial-scale=1.0"> <title>IR PLANNER</title> <link rel="stylesheet" href="Styles\general_styles.css"> <link rel="stylesheet" href="Styles\citymanager_styles.css"> </head> <body> <form class="modal" action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST"> <label for="city">City:</label> <input type="text" name="city" id="city"> <label for="from_station">From Station:</label> <input type="text" name="from_station" id="from_station"> <label for="from_date">From Date:</label> <input type="date" name="from_date" id="from_date"> <input type="submit" value="Submit" id="Submit"> </form> </body> </html> 按下提交按钮时，不会显示错误。然而，当检查数据库表时，什么也没有出现。我已阅读文章，并且 sql 连接（config.php）工作正常（在其他页面和文件中工作）并且 sql 查询也很好。有人遇到这个问题吗？您需要在提交按钮中传递 name 属性，直到您无法从中获取任何值。在您的条件下，它总是返回 false。所以不会更新任何内容，也不会生成错误。只需更新这里是更新的代码。 <?php session_start(); include('config.php'); if(isset($_POST['Submit'])) { $city = $_POST['city']; $from_station = $_POST['from_station']; $from_date = $_POST['from_date']; // Prepare and execute the SQL query $sql = "INSERT INTO journeys (city, from_station, from_date) VALUES ('$city', '$from_station', '$from_date')"; $result = $conn->query($sql); if (mysqli_query($conn, $sql)) { echo "New record created successfully"; } else { echo "Error: " . $sql . "<br>" . mysqli_error($conn); } } ?> <!DOCTYPE html> <html lang="pt_PT"> <head> <meta charset="UTF-8"> <meta name="viewport" content="width=device-width, initial-scale=1.0"> <title>IR PLANNER</title> <link rel="stylesheet" href="Styles\general_styles.css"> <link rel="stylesheet" href="Styles\citymanager_styles.css"> </head> <body> <form class="modal" action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST"> <label for="city">City:</label> <input type="text" name="city" id="city"> <label for="from_station">From Station:</label> <input type="text" name="from_station" id="from_station"> <label for="from_date">From Date:</label> <input type="date" name="from_date" id="from_date"> <input type="submit" name="Submit" value="Submit" id="Submit"> </form> </body> </html> 您可以采取一些措施来分析问题/优化代码。 1.要调试代码：回显一个值以检查是否进入 if 语句例如： if(isset($_POST['Submit'])) { var_dump(__LINE__); 现在运行代码以查看到达了哪些行。 2. if 语句中的代码不会被执行，因为：此代码： <input type="submit" value="Submit" id="Submit"> 需要 name 属性，才能作为数据发送到服务器： <input type="submit" name="Submit" value="Submit" id="Submit"> 3.使用mysqliprepare准备sql语句代替当前的插入代码，制作它 $stmt = $conn->prepare("INSERT INTO journeys (city, from_station, from_date) VALUES (?, ?, ?)"); $stmt->bind_param("sss", $city, $from_station, $from_date); $stmt->execute(); 这样就不会发生SQL注入了。 4.在您的代码中，您还执行了两次 SQL。在这些线上： $result = $conn->query($sql); if (mysqli_query($conn, $sql)) { 替换为第 3 点的代码。生成的代码 <?php session_start(); include('config.php'); if(isset($_POST['Submit'])) { $city = $_POST['city']; $from_station = $_POST['from_station']; $from_date = $_POST['from_date']; // Prepare and execute the SQL query $stmt = $conn->prepare("INSERT INTO journeys (city, from_station, from_date) VALUES (?, ?, ?)"); $stmt->bind_param("sss", $city, $from_station, $from_date); $stmt->execute(); } ?> <!DOCTYPE html> <html lang="pt_PT"> <head> <meta charset="UTF-8"> <meta name="viewport" content="width=device-width, initial-scale=1.0"> <title>IR PLANNER</title> <link rel="stylesheet" href="Styles\general_styles.css"> <link rel="stylesheet" href="Styles\citymanager_styles.css"> </head> <body> <form class="modal" action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST"> <label for="city">City:</label> <input type="text" name="city" id="city"> <label for="from_station">From Station:</label> <input type="text" name="from_station" id="from_station"> <label for="from_date">From Date:</label> <input type="date" name="from_date" id="from_date"> <input type="submit" name="Submit" value="Submit" id="Submit"> </form> </body> </html>

php mysql mysqli sql-insert

回答 2 投票 0

在PHP中执行SQL查询时，没有显示错误，但数据库中看不到数据。我应该检查什么？

php mysql mysqli sql-insert

回答 2 投票 0

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