無法弄清楚爲什麼我的PHP沒有插入到MySQL DB中

Question

我試圖從簡單的註冊表格中插入數據到MySQL中。我有我的數據庫託管在亞馬遜AWS RDS上，使用DBeaver編輯它。當我運行代碼時，我得到以下內容

affected_rows。「插入到數據庫中的數據」。 } else {echo「發生了 錯誤，未添加項目。」; } $ db-> close（）; ？>

我該如何解決這個問題？我的PHP錯了嗎？我很疑惑我是否可以使用MySQLi，我將如何確定？我假設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(); 
?> 

這裏是從HTML頁面的形式

<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

你的代碼是在SQL注入風險，你應該使用PARAM結合，而不是var當成$ _ POST 反正尊重你質疑失蹤插入值 可能因爲涉及到你指的$ _ POST指數在錯誤的方法其實

例如：使用級聯，你應該

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