PHP MySQLi在表中插入的行太多

我需要使用PHP和MySQLi将HTML表单中的数据插入数据库

但是我的代码工作不正常。它将相同数据的几行插入表中

我到处寻找答案，但没有任何帮助。我是新手，所以请帮我找出问题所在。如果有更好的方法做某事，我也希望看到

多谢各位

connect.php：

$link = mysqli_connect("localhost", "root", "", "tutorial");
// Check connection
if ($link === false) {
    die("ERROR: Could not connect. " . mysqli_connect_error());
}
else {
    // echo "Connected. ";
}

<?php
    // Connection
    include("connect.php");
?>

<form action="insert.php" method="post">
    <p>
        <label for="firstName">First Name:</label>
        <input type="text" name="firstname" id="firstName">
    </p>
    <p>
        <label for="lastName">Last Name:</label>
        <input type="text" name="lastname" id="lastName">
    </p>
    <p>
        <label for="Country">Country:</label>
        <select name="country_id" id="Country">
            <?php
                $sql = mysqli_query($link, "SELECT * FROM country");
                while ($row = $sql->fetch_assoc()) {
                    unset($country, $name);
                    $country = $row['country_id'];
                    $name = $row['name'];
                    echo '<option value="'.$country.'">'.$name.'</option>';
                }
            ?>
        </select>
    </p>
    <input type="submit" name="submit" value="Submit">
</form>

<?php
    // Connection
    include("connect.php");

    // Prepare an insert statement
    $sql = "INSERT INTO user (country_id, firstname, lastname) VALUES (?,?, ?)";

    if ($stmt = mysqli_prepare($link, $sql)) {
        // Bind variables to the prepared statement as parameters
        mysqli_stmt_bind_param($stmt, "sss", $country, $firstname, $lastname);

        // Set the parameters
        $country = $_REQUEST['country_id'];
        $firstname = $_REQUEST['firstname'];
        $lastname = $_REQUEST['lastname'];
        mysqli_stmt_execute($stmt);

        // Attempt to execute the prepared statement
        if (mysqli_stmt_execute($stmt)) {
            echo "Records inserted successfully.";
        }
        else {
            echo "ERROR: Could not execute query: $sql. " . mysqli_error($link);
        }
    }
    else {
        echo "ERROR: Could not prepare query: $sql. " . mysqli_error($link);
    }

    // Close statement
    mysqli_stmt_close($stmt);

    // Close connection
    mysqli_close($link);
?>

form.php：

$link = mysqli_connect("localhost", "root", "", "tutorial");
// Check connection
if ($link === false) {
    die("ERROR: Could not connect. " . mysqli_connect_error());
}
else {
    // echo "Connected. ";
}

<?php
    // Connection
    include("connect.php");
?>

<form action="insert.php" method="post">
    <p>
        <label for="firstName">First Name:</label>
        <input type="text" name="firstname" id="firstName">
    </p>
    <p>
        <label for="lastName">Last Name:</label>
        <input type="text" name="lastname" id="lastName">
    </p>
    <p>
        <label for="Country">Country:</label>
        <select name="country_id" id="Country">
            <?php
                $sql = mysqli_query($link, "SELECT * FROM country");
                while ($row = $sql->fetch_assoc()) {
                    unset($country, $name);
                    $country = $row['country_id'];
                    $name = $row['name'];
                    echo '<option value="'.$country.'">'.$name.'</option>';
                }
            ?>
        </select>
    </p>
    <input type="submit" name="submit" value="Submit">
</form>

<?php
    // Connection
    include("connect.php");

    // Prepare an insert statement
    $sql = "INSERT INTO user (country_id, firstname, lastname) VALUES (?,?, ?)";

    if ($stmt = mysqli_prepare($link, $sql)) {
        // Bind variables to the prepared statement as parameters
        mysqli_stmt_bind_param($stmt, "sss", $country, $firstname, $lastname);

        // Set the parameters
        $country = $_REQUEST['country_id'];
        $firstname = $_REQUEST['firstname'];
        $lastname = $_REQUEST['lastname'];
        mysqli_stmt_execute($stmt);

        // Attempt to execute the prepared statement
        if (mysqli_stmt_execute($stmt)) {
            echo "Records inserted successfully.";
        }
        else {
            echo "ERROR: Could not execute query: $sql. " . mysqli_error($link);
        }
    }
    else {
        echo "ERROR: Could not prepare query: $sql. " . mysqli_error($link);
    }

    // Close statement
    mysqli_stmt_close($stmt);

    // Close connection
    mysqli_close($link);
?>

名字：


姓氏：


国家：

insert.php：

$link = mysqli_connect("localhost", "root", "", "tutorial");
// Check connection
if ($link === false) {
    die("ERROR: Could not connect. " . mysqli_connect_error());
}
else {
    // echo "Connected. ";
}

<?php
    // Connection
    include("connect.php");
?>

<form action="insert.php" method="post">
    <p>
        <label for="firstName">First Name:</label>
        <input type="text" name="firstname" id="firstName">
    </p>
    <p>
        <label for="lastName">Last Name:</label>
        <input type="text" name="lastname" id="lastName">
    </p>
    <p>
        <label for="Country">Country:</label>
        <select name="country_id" id="Country">
            <?php
                $sql = mysqli_query($link, "SELECT * FROM country");
                while ($row = $sql->fetch_assoc()) {
                    unset($country, $name);
                    $country = $row['country_id'];
                    $name = $row['name'];
                    echo '<option value="'.$country.'">'.$name.'</option>';
                }
            ?>
        </select>
    </p>
    <input type="submit" name="submit" value="Submit">
</form>

<?php
    // Connection
    include("connect.php");

    // Prepare an insert statement
    $sql = "INSERT INTO user (country_id, firstname, lastname) VALUES (?,?, ?)";

    if ($stmt = mysqli_prepare($link, $sql)) {
        // Bind variables to the prepared statement as parameters
        mysqli_stmt_bind_param($stmt, "sss", $country, $firstname, $lastname);

        // Set the parameters
        $country = $_REQUEST['country_id'];
        $firstname = $_REQUEST['firstname'];
        $lastname = $_REQUEST['lastname'];
        mysqli_stmt_execute($stmt);

        // Attempt to execute the prepared statement
        if (mysqli_stmt_execute($stmt)) {
            echo "Records inserted successfully.";
        }
        else {
            echo "ERROR: Could not execute query: $sql. " . mysqli_error($link);
        }
    }
    else {
        echo "ERROR: Could not prepare query: $sql. " . mysqli_error($link);
    }

    // Close statement
    mysqli_stmt_close($stmt);

    // Close connection
    mysqli_close($link);
?>

您正在调用mysqli\u stmt\u execute（$stmt）两次

不确定是否有帮助，但请尝试更改此设置：

1 - mysqli_stmt_execute($stmt);
// Attempt to execute the prepared statement
2 - if(mysqli_stmt_execute($stmt)){
    echo "Records inserted successfully.";
} 

---

您可以删除数字1。

mysqli\u stmt\u执行

返回true或false，将返回的值分配给var（更多信息）

替换这个

mysqli_stmt_execute($stmt);

// Attempt to execute the prepared statement
if(mysqli_stmt_execute($stmt)){
    echo "Records inserted successfully.";
}

用这个

$execution = mysqli_stmt_execute($stmt);

// Attempt to execute the prepared statement
if($execution){
    echo "Records inserted successfully.";
}

---

您正在调用mysqli_stmt_execute（）2两次

下面是如何用面向对象的风格来实现它

connect.php：

<?php
    $mysqli = new Mysqli("localhost", "root", "", "tutorial");
    /* check connection */
    if ($mysqli->connect_errno) {
        printf("Connect failed: %s\n", $mysqli->connect_error);
        exit();
    }
?>

insert.php：

<?php
    // Connection
    require_once "connect.php";

    $query = "INSERT INTO user (country_id, firstname, lastname) VALUES (?,?, ?)";
    if($stmt = $mysqli->prepare($query)) {

        $country = $_REQUEST['country_id'];
        $firstname = $_REQUEST['firstname'];
        $lastname = $_REQUEST['lastname'];

        $stmt->bind_param('sss',$country,$firstname,$lastname);

        if($stmt->execute()) {

            echo 'Success ! ';

            $stmt->close();
            $mysqli->close();

        } else {
            echo "Could not execute query: " . $mysqli->error();
            $stmt->close();
            $mysqli->close();
        }

    } else {
       echo 'Error preparing statement : ' . $mysqli->error;
       $stmt->close();
       $mysqli->close();
    }

---

欢迎来到Stackoverflow！请花点时间看看如何提出好的问题。这也将帮助你得到最好的答案。（）非常感谢，使用面向对象的方式更好吗？您需要保持代码的干净性和可理解性。最好的方法是使用oop，这样您就可以将类似的任务分组到类中，并消除重复代码的需要。