Question

客户将填写表格并输入他们希望将CSV导出到的路径。使用php的顶部（下面）输入该路径：

    <form action="register.php" method="post">
        Enter file pathway where CSV will be saved: <input type="text" name="username" required="required"/> <br/>
        <input type="submit" value="Enter"/>
    </form>
</body>

我想创建一个名为path的变量。目前我可以将文本输入到mysql数据库中的正确行（我可以将John打印在数据库中），但不是输入到表单中的正确文本（即$途径）。我想创建一个变量，因为在数据库中保存路径之后我还想在export.php中使用它。

我假设我也需要这样的东西：

if($_SERVER["REQUEST_METHOD"] == "POST"){
    $pathway = mysql_real_escape_string($_POST['pathway']); 
// but i can't seem to piece it altogether.


    <?php
    $servername = "localhost";
    $username = "";
    $password = "";
    $dbname = "first_db";
    $table_users = $row['pathway'];
    $pathway = "pathway";                                             

    // Create connection
    $conn = mysqli_connect($servername, $username, $password, $dbname);
    // Check connection
    if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
    }

    $sql = "INSERT INTO users (pathway)
    VALUES ('John')";

    if (mysqli_query($conn, $sql)) {
    echo "New record created successfully";
    } else {
    echo "Error: " . $sql . "<br>" . mysqli_error($conn);
    }

    mysqli_close($conn);
    ?>

Answer 1

如果不是这样，请检查您的用户名和密码......

<?php
$servername = "localhost";
$username = "";
$password = "";
$dbname = "first_db";
$pathway = $_POST['username'];  username - is the name of your input.                                             

// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn)
{
    die("Connection failed: " . mysqli_connect_error());
}

$sql = "INSERT INTO users (pathway)
    VALUES ('$pathway')";

if (mysqli_query($conn, $sql)) {
    echo "New record created successfully";
}
else
{
echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}

mysqli_close($conn);
?>

Answer 2

您的数据库用户名为var input = document.getElementById('id');

将Null更改为$username = "";

Answer 3

您的输入字段名称是用户名

将pathway更改为$_POST['pathway']才能正常工作

<form action="register.php" method="post">
    Enter file pathway where CSV will be saved: 
     <input type="text" name="pathway" required="required"/> <br/>
    <input type="submit" value="Enter"/>
</form>

Answer 4

首先，您已获得用户名＆＃39;作为用于输入路径的字段的名称，因此将其重命名为＆＃39;途径＆＃39;。我不知道我是否了解你，但你只是想阅读发布的内容吗？尝试类似：

$pathway = $_POST['pathway']

我强烈建议使用面向对象的样式

$conn = new mysqli...

然后

mysqli->prepare(), mysqli->bind_param(), mysqli->execute()

有了这个，你不必处理mysqli_real_escape_string等。

使用php

4 个答案: