Question

我试图以两种方式做到这一点。我在html和php上使用bootstrap将数据发送到数据库。在发布之前，我仔细阅读stackoverflow并尝试在谷歌中找到的所有内容。请帮忙，我是php的新手并且真的想学习它，但这个麻烦让我在这一刻超过3个小时。抱歉我的英语不好（对于这种情况，我不是来自印度）第一个变种。 HTML：

    <!DOCTYPE html>
<html lang="en">
    <head>
        <meta http-equiv="content-type" content="text/html; charset=UTF-8">
        <meta charset="utf-8">
        <title>Landing Page template</title>
        <meta name="generator" content="Bootply" />
        <meta name="viewport" content="width=device-width, initial-scale=1, maximum-scale=1">
        <link href="css/bootstrap.min.css" rel="stylesheet">
        <link href="//netdna.bootstrapcdn.com/font-awesome/3.2.1/css/font-awesome.min.css" rel="stylesheet">
        <!--[if lt IE 9]>
            <script src="//html5shim.googlecode.com/svn/trunk/html5.js"></script>
        <![endif]-->
        <link href="css/styles.css" rel="stylesheet">
    </head>
    <body>
    <div class="background-image"></div>
<div class="container-full">

      <div class="row">

        <div class="col-lg-12 text-center v-center">

          <h1>Hello</h1>
          <p class="lead"><u>Sign-up and receive.</u></p>
          <p class="lead">No ads, no paid scum. No Bullshit.</p>
          <br><br><br>

          <form class="col-lg-12" action="dbconnection.php" method="POST">
            <div class="input-group" style="width:340px;text-align:center;margin:0 auto;">
            <input class="form-control input-lg" title="Don't worry. We hate spam, and will not share your email with anyone." placeholder="Your email" type="text" name="name">
              <span class="input-group-btn" >
              <button class="btn btn-lg btn-primary"  type="submit"  
      name="submit" value=" Send">SUBMIT</button></span>

            </div>
          </form>
        </div>

      </div> 
      <div class="row text-center v-center">
       <p class="lead">We Deliver: </p>


      </div>
      <!-- /row -->





    <!-- script references -->
        <script src="//ajax.googleapis.com/ajax/libs/jquery/2.0.2/jquery.min.js"></script>
        <script src="js/bootstrap.min.js"></script>
    </body>
</html>

PHP：

<?php
// connect to the database
$user_name = "root";
$password = "";
$database = "emails";
$host_name = "localhost";
$name = $_POST["name"];
$con = mysqli_connect ("localhost","root","","emails") or die("Error ".mysqli_error($con));
//check connection
$name222 = $_POST["name"];
$name=mysqli_real_escape_string($_POST['name']);
mysqli_query("INSERT INTO emails (email) VALUES ('$name')");
echo"record added";
$result = mysqli_query();
if($result){

    echo("<br>Input data is succeed");

} else{

    echo("<br>Input data is fail");

}
echo "Connection opened";

mysql_close($con); 
?>

错误：

警告：mysqli_real_escape_string（）只需要2个参数，1   在第11行的C：\ OpenServer \ domains \ landing-cpa \ dbconnection.php中给出

警告：mysqli_query（）需要至少2个参数，1中给出1   第12行记录中的C：\ OpenServer \ domains \ landing-cpa \ dbconnection.php   添加警告：mysqli_query（）需要至少2个参数，给定0   在第14行的C：\ OpenServer \ domains \ landing-cpa \ dbconnection.php

输入数据是failConnection打开警告：mysql_close（）期望   参数1是资源，给定的对象   第26行的C：\ OpenServer \ domains \ landing-cpa \ dbconnection.php

第二种变体： HTML：相同 PHP：

 <?php
$servername = "localhost";
$username = "root";
$password = "";
$database = "emails";

// Create connection
$conn = new mysqli($servername, $username, $password, $database);
mysql_select_db("$database");

$name = $_POST["name"];
$order = "INSERT INTO emails

        (email)

        VALUES

        ('$name')";


$result = mysql_query($order);

if($result){

    echo("<br>Input data is succeed");

} else{

    echo("<br>Input data is fail");

}


// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}
echo "Connected successfully";
?>

错误：

警告：mysql_select_db（）：拒绝用户访问＆＃39;＆＃39; localhost＆＃39;   （使用密码：NO）in   第12行的C：\ OpenServer \ domains \ landing-cpa \ dbconnection.php

警告：mysql_select_db（）：无法链接到服务器   建立在C：\ OpenServer \ domains \ landing-cpa \ dbconnection.php上   第12行

警告：mysql_query（）：拒绝用户＆＃39;＆＃39; localhost＆＃39; （使用   密码：NO）在C：\ OpenServer \ domains \ landing-cpa \ dbconnection.php上   第24行

警告：mysql_query（）：无法建立指向服务器的链接   在第24行的C：\ OpenServer \ domains \ landing-cpa \ dbconnection.php

输入数据已成功连接

Answer 1

我已经这样做了，只需在这里阅读mysqli_real_escape_string http://www.w3schools.com/php/func_mysqli_real_escape_string.asp

获得有效的解决方案主要错误在于我的语法（mysql_string命令），下面的代码完全有用，mb它将帮助未来的人：

S::Foo

PHP ERROR无法将数据从表单发送到mysql

1 个答案: