注册表单PHP SQL
我正在做一份登记表。单击提交时，此表单需要向数据库添加数据。它没有给出任何错误或消息。这是我的代码，希望有人能帮我注册表单PHP SQL,php,mysqli,Php,Mysqli,我正在做一份登记表。单击提交时，此表单需要向数据库添加数据。它没有给出任何错误或消息。这是我的代码，希望有人能帮我 <body> <form action="" method"post" class="form"> <div class="form-group1"> <label for="" >Username :</label> <input type="text" class="form-contr
<body>
<form action="" method"post" class="form">
<div class="form-group1">
<label for="" >Username :</label>
<input type="text" class="form-control" name="username" id="tb-username" placeholder="Username">
</div>
<div class="form-group2">
<label for="" >Password :</label>
<input type="password" class="form-control" name="password" id="tb-password" placeholder="Password">
</div>
<button type="submit" class="btn btn-primary" id="btn-submit" name="register">Submit</button>
</form>
<?php
if(isset($_POST['register']))
{
if(isset($_POST['username'], $_POST['password']))
{
$username = $_POST['username'];
$password = $_POST['password'];
$link = mysqli_connect("localhost", "root", " ", "vbproject");
if(mysqli_connect_errno())
{
printf("Connect failed: %s\n", mysqli_connect_errno());
exit();
}
$sql = "INSERT INTO users (username, password)
VALUES('$username', '$password')";
mysqli_close($link);
}
}
?>
</body>
用户名：
密码：
提交
这就是流动的方式
<?php
$con=mysqli_connect("localhost", "root", " ", "vbproject");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
// Perform queries
$sql = "INSERT INTO users (username, password)
VALUES('$username', '$password')";
mysqli_query($con, $sql);
mysqli_close($con);
?>
在关闭连接之前添加此行，它将执行您没有执行的查询
mysqli_query($link,$sql);
这里有个错误
<form action="" method"post" class="form">
替换为
<form action="" method="post" class="form">
您错过了=
标志，导致您的请求为GET
请求，您正在以POST
方法访问您也需要：
$sql = "INSERT INTO users (username, password)
VALUES('$username', '$password')";
$result = $link->query($sql);
if($result === true){
echo "Insert Success";
}else{
echo "Insert failed";
}
请将method“post”
更改为method=“post”
格式标签
<body>
<form action="" method="post" class="form">
<div class="form-group1">
<label for="" >Username :</label>
<input type="text" class="form-control" name="username" id="tb-username" placeholder="Username">
</div>
<div class="form-group2">
<label for="" >Password :</label>
<input type="password" class="form-control" name="password" id="tb-password" placeholder="Password">
</div>
<button type="submit" class="btn btn-primary" id="btn-submit" name="register">Submit</button>
</form>
<?php
if(isset($_POST['register']))
{
if(isset($_POST['username'], $_POST['password']))
{
$username = $_POST['username'];
$password = $_POST['password'];
$link = mysqli_connect("localhost", "root", " ", "vbproject");
if(mysqli_connect_errno())
{
printf("Connect failed: %s\n", mysqli_connect_errno());
exit();
}
$sql = "INSERT INTO users (username, password)
VALUES('$username', '$password')";
if(mysqli_query($link, $sql)){
echo 'Inserted';
}else{
echo 'Not Inserted';
}
mysqli_close($link);
}
}
?>
</body>
您没有运行mysql查询。请找到以下解决方案：
$sql = "INSERT INTO users (username, password)
VALUES('$username', '$password')";
mysqli_query($con, $sql); // this Line Missing
mysqli_close($link);
只是观察一下你的代码
<?php
if(isset($_POST['register']))
{
您不准备或执行SQL。存储普通密码，mmmmmmysqli_查询（$con，$SQL）；//SQL查询后缺少这一行。@u_mulder（到目前为止）的一个答案解决了存储普通密码的问题，这在当今的安全性方面非常重要。@Rick_Jellema您不需要给出只解决一个问题的答案，但答案应该显示解决问题的最佳方法。
<?php
if(isset($_POST['register']))
{
<button type="submit" class="btn btn-primary" id="btn-submit" name="register">Submit</button>