错误：未选择数据库且未定义索引

Question

我在这里卡了将近一天，我只是找不到正确的解决方案，请帮助我收到未选择数据库的错误，但我收到了连接已创建的消息 我得到的第二个错误是 undefined index "SNo" "clientname" "contact no" "option-3 (its a checkbox)"

HTML格式

<form  action="survey-data-entry.php" name="1st-form" id="form-control" onSubmit="return validate()" method="post" >    
    <label>S.No.</label>
<input type="text" class="form-control" placeholder=" 0451211315" name="SNo" id="s.no"  >
<label>Client Name</label>
<input type="text" class="form-control" name="clientname" id="clientname" />                
<label>Contact No</label>
<input type="text" class="form-control" name="contactno" id="contactno" />
    <strong>Pest Buster  </strong>
<input type="checkbox" name="option_3" id="option_3" value="Pest Buster">

这是 PHP 代码

$hostname = 'localhost';
$username ='root';
$password = '';
$dbname = 'survey-form';

$conn = mysql_connect($hostname, $username, $password, $dbname);
if(!$conn){
    die("connection failed:" .mysql_error());
}
echo "connection hogya successful";



$SNo = $_POST['SNo'];
     $clientname = $_POST['clientname'];
     $contactno = $_POST['contactno'];
     $option_3 = $_POST['option_3'];
$sql= "INSERT INTO php-startup (SNo, clientname, contactno, option_3)
VALUES('SNo', 'clientname', 'contactno ','option_3')";

if(!mysql_query($sql))
{
die('caution' . mysql_error());
}
echo"1 record added";

Answer 1

我不知道“未选择数据库”，因为我没有使用mysql_*的经验

你可以先检查变量是否设置，像这样

if(isset($_POST['SNo']))
{
    $SNo = $_POST['SNo'];
    $clientname = $_POST['clientname'];
    $contactno = $_POST['contactno'];
    $option_3 = $_POST['option_3'];
    $sql= "INSERT INTO php-startup (SNo, clientname, contactno, option_3)VALUES('SNo', 'clientname', 'contactno ','option_3')";

    if(!mysql_query($sql))
    {
        die('caution' . mysql_error());
    }
    echo"1 record added";
}

这将删除未定义的索引错误。并停止使用mysql_* 语法，它们是depecrated。相反，学习使用pdo 或mysqli

Answer 2

数据库连接

$connection = mysql_connect('localhost', 'mysql_user', 'mysql_password');

然后选择db

mysql_select_db('database_name',$connection);

然后执行查询

if(isset($_POST['SNo'])){
         $SNo = $_POST['SNo'];
         $clientname = $_POST['clientname'];
         $contactno = $_POST['contactno'];
         $option_3 = $_POST['option_3'];
    $sql= "INSERT INTO php-startup (SNo, clientname, contactno, option_3)
    VALUES('$SNo', '$clientname', '$contactno','$option_3')";

    if(!mysql_query($sql))
    {
        die('caution' . mysql_error());
    }
        echo"1 record added";
}

Answer 3

mysql_connect 不将数据库名称作为参数see docs。 改变->

mysql_connect($hostname, $username, $password, $dbname);

->

 $con = mysql_connect($host, $user, $pass);
    if (!$con) {
        die('Could not connect: ' . mysql_error());
    }
    echo 'Connected successfully';
    mysql_select_db($database_name, $con)
    mysql_close($con);

我会建议使用 mysqli_connect see documentation。

$link = mysqli_connect($host, $user, $password, $database_name);

if (!$link) {
    echo "Error: Unable to connect to MySQL." . PHP_EOL;
}