使用php表单将值存储到数据库中
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【中文标题】使用php表单将值存储到数据库中【英文标题】：Storing the values into database using php forms 【发布时间】：2016-01-21 15:09:00 【问题描述】：最终编辑： 谢谢大家的帮助。我一直在尝试在 index.php 而不是 submit.php 中编写与连接相关的所有代码。现在已经解决了。
编辑：
我已根据您的反馈更新了代码。 我现在可以将值获取到数据库，但问题是它只显示空结果。这是更新的代码。
<form action="submit.php" method="post" class="form-horizontal">
<div class="form-group">
<label for="name" class ="col-lg-2 control-label" > Name</label>
<div class="col-lg-7">
<input type="text" class="form-control" id ="name" name="name" placeholder="Enter your Name" required>
</div>
</div>
</div>
<div class="col-lg-1">
</div>
<div class="form-horizontal" >
<div class="form-group">
<label for="email" class ="col-lg-2 control-label" > Email</label>
<div class="col-lg-7">
<input type="text" class="form-control" id ="email" name="email" placeholder="Enter your email address" required>
</div>
</div>
</div> <div class="col-lg-1">
</div>
<div class="form-horizontal" >
<div class="form-group">
<label for="subject" class ="col-lg-2 control-label" > Subject</label>
<div class="col-lg-7">
<input type="text" class="form-control" id ="subject" name="subject" placeholder="Your Subject" required>
</div>
</div>
</div>
<div class="col-lg-1">
</div>
<div class="form-horizontal">
<div class="form-group">
<label for="message" class ="col-lg-2 control-label" > Message</label>
<div class="col-lg-7">
<textarea name="message" class="form-control" id ="message" cols="20" rows="3" placeholder="Your Message"></textarea>
</div>
</div> <!-- end form -->
<div class="col-lg-1">
</div>
<div class="form-group">
<div class="col-lg-7 col-lg-offset-2">
<button type="submit" name="submit" class="btn btn-primary">Submit</button>
</div>
</div>
</form>
PHP 代码：
<?php
if (isset($_POST))
$conn = mysqli_connect($servername, $username, $password, $db_name);// Establishing Connection with Server
mysqli_set_charset($conn, 'utf8');
if (!$conn)
die("Database connection failed: " . mysqli_error($conn));
else
echo "connected successfully";
//Escaping string, not 100% safe, also consider validating rules and sanitization
$name = mysqli_real_escape_string($conn, $_POST['name']);
$email = mysqli_real_escape_string($conn, $_POST['email']);
$subject = mysqli_real_escape_string($conn, $_POST['subject']);
$message = mysqli_real_escape_string($conn, $_POST['message']);
$result = mysqli_query($conn, "INSERT INTO contact (user, email, subject, message) VALUES ('$name', '$email', '$subject', '$message')");
?>
Here is the snapshot of the database
我有一个使用 html 制作的表单。当我在数据库中提交表单时，我想存储结果。连接成功，但数据未存储在数据库中。
submit.php 所做的基本上只是发送文本“成功提交表单”。
这是我的代码：
<form action="submit.php" method="post" class="form-horizontal">
<div class="form-group">
<label for="name" class ="col-lg-2 control-label" > Name</label>
<div class="col-lg-7">
<input type="text" class="form-control" id ="name" placeholder="Enter your Name" required>
</div>
</div>
</div>
<div class="col-lg-1">
</div>
<div class="form-horizontal" >
<div class="form-group">
<label for="email" class ="col-lg-2 control-label" > Email</label>
<div class="col-lg-7">
<input type="text" class="form-control" id ="email" placeholder="Enter your email address" required>
</div>
</div>
</div> <div class="col-lg-1">
</div>
<div class="form-horizontal" >
<div class="form-group">
<label for="subject" class ="col-lg-2 control-label" > Subject</label>
<div class="col-lg-7">
<input type="text" class="form-control" id ="subject" placeholder="Your Subject" required>
</div>
</div>
</div>
<div class="col-lg-1">
</div>
<div class="form-horizontal">
<div class="form-group">
<label for="message" class ="col-lg-2 control-label" > Message</label>
<div class="col-lg-7">
<textarea name="message" class="form-control" id ="message" cols="20" rows="3" placeholder="Your Message"></textarea>
</div>
</div> <!-- end form -->
<div class="col-lg-1">
</div>
<div class="form-group">
<div class="col-lg-7 col-lg-offset-2">
<button type="submit" class="btn btn-primary">Submit</button>
</div>
</div>
</form>
PHP 代码：
$conn = mysqli_connect($servername, $username, $password, $db_name);// Establishing Connection with Server
mysqli_set_charset($conn, 'utf8');
if (!$conn)
die("Database connection failed: " . mysqli_error($conn));
else
echo "connected successfully";
if (isset($_POST['submit']))
//Escaping string, not 100% safe, also consider validating rules and sanitization
$name = mysqli_real_escape_string($conn, $_POST['name']);
$email = mysqli_real_escape_string($conn, $_POST['email']);
$subject = mysqli_real_escape_string($conn, $_POST['subject']);
$message = mysqli_real_escape_string($conn, $_POST['message']);
$result = mysqli_query($conn, "INSERT INTO contact (user, email, subject, message) VALUES ('$name', '$email', '$subject', '$message');");
if ($result)
$message="successfully sent the query!!";
else
$message="try again!!";
?>
【问题讨论】：
【参考方案1】：您的所有输入字段都没有name="" 属性，包括按钮。所以这些字段都不会在 $_POST 数组中发送。
将这样的name="" 属性添加到要发送到 PHP 的所有字段
<form action="submit.php" method="post" class="form-horizontal">
<div class="form-group">
<label for="name" class ="col-lg-2 control-label" > Name</label>
<div class="col-lg-7">
<input type="text" class="form-control" id ="name" name="name" placeholder="Enter your Name" required>
</div>
</div>
</div>
<div class="col-lg-1">
</div>
<div class="form-horizontal" >
<div class="form-group">
<label for="email" class ="col-lg-2 control-label" > Email</label>
<div class="col-lg-7">
<input type="text" class="form-control" id ="email" name="email" placeholder="Enter your email address" required>
</div>
</div>
</div>
<div class="col-lg-1"></div>
<div class="form-horizontal" >
<div class="form-group">
<label for="subject" class ="col-lg-2 control-label" > Subject</label>
<div class="col-lg-7">
<input type="text" class="form-control" id ="subject" name="subject" placeholder="Your Subject" required>
</div>
</div>
</div>
<div class="col-lg-1"></div>
<div class="form-horizontal">
<div class="form-group">
<label for="message" class ="col-lg-2 control-label" > Message</label>
<div class="col-lg-7">
<textarea name="message" class="form-control" id ="message" name="message" cols="20" rows="3" placeholder="Your Message"></textarea>
</div>
</div> <!-- end form -->
<div class="col-lg-1"></div>
<div class="form-group">
<div class="col-lg-7 col-lg-offset-2">
<button type="submit" class="btn name="submit" btn-primary">Submit</button>
</div>
</div>
</form>
同样在submit.php 中的代码中更改此设置，以便在发生错误消息时看到实际的错误消息。
if ($result)
$message="successfully sent the query!!";
else
$message="Insert failed : " . mysqli_error($conn);
echo $message;
虽然这确实假设您实际上在代码中的某处显示了 $message 值，但您没有向我们显示。
【讨论】：
我已经添加了名称字段，但它仍然是一样的。 if($result) 会发生什么？我没有收到任何消息。一旦我点击提交，它就会进入 submit.php 页面 好的，也请参阅为您的数据库访问错误处理建议的 mod 您能看一下更新后的代码吗？它正在返回空字段 你没有修改查询错误检查，实际上看起来你完全删除了它 是的，我做到了。它正在连接到数据库，每当我提交表单时，都会存储空值。我认为可能有一个小问题。我真的很感激你能谈谈看看它。【参考方案2】：您必须将name 属性添加到您的button 元素，以便if (isset($_POST['submit'])) 将是true。
请更改
<button type="submit" class="btn btn-primary">Submit</button>
到
<button type="submit" name="submit" class="btn btn-primary">Submit</button>
或
<input type="submit" name="submit" value="Submit" class="btn btn-primary" />
【讨论】：
if (isset($_POST['submit'])) ... 里面的代码到达了吗？也请在您的mysqli_query 之后添加var_dump($result); 以检查您的查询结果。
那是因为此规则适用于所有输入字段。你的都没有name="" 属性，所以不会在 POST 上发送
我添加了一个名称字段，但它仍然是一样的。
您必须指定哪些代码被执行，哪些代码没有被执行，以便我们定位问题。
我有一些更改@mapek。检查编辑并让我知道问题所在【参考方案3】：
首先你必须为每个输入标签和按钮标签提供 name 属性以获得更好的方法：
<form action="submit.php" method="post" class="form-horizontal">
<div class="form-group">
<label for="name" class ="col-lg-2 control-label" > Name</label>
<div class="col-lg-7">
<input type="text" class="form-control" id ="name" name ="name" placeholder="Enter your Name" required>
</div>
</div>
</div>
<div class="col-lg-1">
</div>
<div class="form-horizontal" >
<div class="form-group">
<label for="email" class ="col-lg-2 control-label" > Email</label>
<div class="col-lg-7">
<input type="text" class="form-control" name ="email" id ="email" placeholder="Enter your email address" required>
</div>
</div>
</div> <div class="col-lg-1">
</div>
<div class="form-horizontal" >
<div class="form-group">
<label for="subject" class ="col-lg-2 control-label" > Subject</label>
<div class="col-lg-7">
<input type="text" class="form-control" name ="subject" id ="subject" placeholder="Your Subject" required>
</div>
</div>
</div>
<div class="col-lg-1">
</div>
<div class="form-horizontal">
<div class="form-group">
<label for="message" class ="col-lg-2 control-label" > Message</label>
<div class="col-lg-7">
<textarea name="message" class="form-control" name ="message" id ="message" cols="20" rows="3" placeholder="Your Message"></textarea>
</div>
</div> <!-- end form -->
<div class="col-lg-1">
</div>
<div class="form-group">
<div class="col-lg-7 col-lg-offset-2">
<button type="submit" name ="submit" class="btn btn-primary">Submit</button>
</div>
</div>
</form>
在数据库中插入数据的 PHP 代码：
$result = mysqli_query($conn, "INSERT INTO contact (user, email, subject, message) VALUES ('".$name."', '".$email."', '".$subject."', '".$message."')");
【讨论】：
我已经修改了代码。请检查编辑并让我知道问题出在哪里。它在数据库中返回空值。也检查数据库图像。 你有没有像我的查询一样在插入查询中连接 php 变量，因为在某些情况下会出现问题... 也试过你的代码。一样的。它现在正在向数据库发送空字段。【参考方案4】：试试这个代码：-
<?php
if ($_SERVER['REQUEST_METHOD'] == 'POST')
$conn = mysqli_connect($servername, $username, $password, $db_name);// Establishing Connection with Server
mysqli_set_charset($conn, 'utf8');
if (!$conn)
die("Database connection failed: " . mysqli_error($conn));
else
echo "connected successfully";
//Escaping string, not 100% safe, also consider validating rules and sanitization
$name = mysqli_real_escape_string($conn, $_POST['name']);
$email = mysqli_real_escape_string($conn, $_POST['email']);
$subject = mysqli_real_escape_string($conn, $_POST['subject']);
$message = mysqli_real_escape_string($conn, $_POST['message']);
$result = mysqli_query($conn, "INSERT INTO contact (user, email, subject, message) VALUES ('$name', '$email', '$subject', '$message')");
echo "INSERT INTO contact (user, email, subject, message) VALUES ('$name', '$email', '$subject', '$message')";//die;
if ($result)
$message="successfully sent the query!!";
else
$message="try again!!";
?>
<form action="index.php" method="post" class="form-horizontal">
<div class="form-group">
<label for="name" class ="col-lg-2 control-label" > Name</label>
<div class="col-lg-7">
<input type="text" class="form-control" name ="name" id ="name" placeholder="Enter your Name" required>
</div>
</div>
</div>
<div class="col-lg-1">
</div>
<div class="form-horizontal" >
<div class="form-group">
<label for="email" class ="col-lg-2 control-label" > Email</label>
<div class="col-lg-7">
<input type="text" class="form-control" id ="email" name="email" placeholder="Enter your email address" required>
</div>
</div>
</div> <div class="col-lg-1">
</div>
<div class="form-horizontal" >
<div class="form-group">
<label for="subject" class ="col-lg-2 control-label" > Subject</label>
<div class="col-lg-7">
<input type="text" name="subject" class="form-control" id ="subject" placeholder="Your Subject" required>
</div>
</div>
</div>
<div class="col-lg-1">
</div>
<div class="form-horizontal">
<div class="form-group">
<label for="message" class ="col-lg-2 control-label" > Message</label>
<div class="col-lg-7">
<textarea name="message" class="form-control" id ="message" cols="20" rows="3" placeholder="Your Message"></textarea>
</div>
</div> <!-- end form -->
<div class="col-lg-1">
</div>
<div class="form-group">
<div class="col-lg-7 col-lg-offset-2">
<button type="submit" class="btn btn-primary" name="submit">Submit</button>
</div>
</div>
</form>
【讨论】：
删除了它。但没用 我已经进行了更改。现在看看。 @Rahautos 不，不是.. 现在，当我提交空字段时，正在存储。检查数据库快照 请再次检查我的答案@user3656329以上是关于使用php表单将值存储到数据库中的主要内容，如果未能解决你的问题，请参考以下文章
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