Question

我试图将数据从简单的寄存器形式插入到MySQL中。我在Amazon AWS RDS上托管了我的数据库，使用DBeaver进行编辑。当我运行代码时，我得到以下

affected_rows＆＃34。数据插入数据库。＆＃34 ;; } else {echo＆＃34; An 错误已经发生。没有添加项目。＆＃34 ;; } $ db-＆gt; close（）; ？＆GT;

我该如何解决这个问题？我的PHP错了吗？我是否可以使用MySQLi感到困惑，我该怎么做？我假设Amazon RDS MySQL兼容。

<?php
// create short variable names
$name=$_POST['name'];
$birthdate=$_POST['birthdate'];
$email=$_POST['email'];
$password=$_POST['password'];
$address=$_POST['address'];
$city+$_POST['city'];

if (!get_magic_quotes_gpc()) {
    $name = addcslashes($name);
    $birthdate = addslashes($birthdate);
    $email = addcslashes($email);
    $password = addcslashes($password);
    $address = addcslashes($address);
    $city = addcslashes($city);
}

$host='xxxxxx'
$user='admin'
$password='xxxxx'
$dbname='users'
@ $db = mysqli_connect($host,$user,$password,$dbname)

if (mysqli_connect_errno()) {
    echo 'Error: Could not connect to database. Please try again later.';
    exit;
}

// Execute the query
$query = "INSERT INTO vestorinfo (name,birthdate,email,password,address,city)
    VALUES ('$_POST[name]','$_POST[birthdate]','$_POST[email]','$_POST[password]','$_POST[address]','$_POST[city]')";
$result = mysqli_query($query)
        or die ("Couldn't execute query."};

if ($result) {
    echo $db->affected_rows." data inserted into database.";
} else {
    echo "An error has occurred. The items were not added.";
}

$db->close();
?>

以下是html页面的表格

    <div id="registerform">
    <form action="php/registerprocess.php" method="post" class="form-horizontal">

<fieldset>

<!-- Form Name -->

<!-- Text input-->
<div class="form-group">
  <label class="col-md-4 control-label" for="name">Name</label>  
  <div class="col-md-4">
  <input id="name" name="name" placeholder="Your name" class="form-control input-md" required="" type="text">

  </div>
</div>

<!-- Text input-->
<div class="form-group">
  <label class="col-md-4 control-label" for="birthdate">Birth Date</label>  
  <div class="col-md-4">
  <input id="birthdate" name="birthdate" placeholder="07/04/1950" class="form-control input-md" required="" type="text">

  </div>
</div>

<!-- Text input-->
<div class="form-group">
  <label class="col-md-4 control-label" for="email">Email</label>  
  <div class="col-md-4">
  <input id="email" name="email" placeholder="" class="form-control input-md" required="" type="text">

  </div>
</div>

<!-- Password input-->
<div class="form-group">
  <label class="col-md-4 control-label" for="password">Password</label>
  <div class="col-md-4">
    <input id="password" name="password" placeholder="" class="form-control input-md" required="" type="password">
    <span class="help-block">Must be &gt;= 8 characters including at least 1 number</span>
  </div>
</div>

<!-- Text input-->
<div class="form-group">
  <label class="col-md-4 control-label" for="email">Address</label>  
  <div class="col-md-4">
  <input id="address" name="address" placeholder="" class="form-control input-md" required="" type="text">
  </div>
</div>

<!-- Text input-->
<div class="form-group">
  <label class="col-md-4 control-label" for="email">City</label>  
  <div class="col-md-4">
  <input id="city" name="city" placeholder="" class="form-control input-md" required="" type="text">
  </div>
</div>
</div>

</fieldset>

Answer 1

你的代码存在sql注入的风险，你应该使用param绑定而不是var作为$ _POST 无论如何，尊重您的问题是缺少插入值可能是因为与你以错误方式引用$ _POST索引的事实有关

例如：使用连接，你应该

  $query = "INSERT INTO vestorinfo (name,birthdate,email,password,address,city)
      VALUES ('" . $_POST['name'] ." , " . $_POST['birthdate'] .", " . 
      $_POST['email'] . "," . $_POST['password'] . "," .
         $_POST['address'] . "," . $_POST['city'] . ")";

无法弄清楚为什么我的PHP不能插入MySQL数据库

1 个答案: