Php 将输入选择选项存储到数据库中

我想用选项的值在数据库中保存输入选择选项。但是没有什么可以挽救。怎么了？我正在数据库中将该值保存为

varchar（10）

表格

<form name="stdntdetails" action="study.php" method="post">
    <select name="department">
        <option value="IT">Information Technology</option>
        <option value="IS">Information System</option>
        <option value="CS">Computer Science</option>
    </select>
    <input type="submit" id="loginbtn" />
</form>

<form name="stdntdetails" action="study.php" method="post">
    <select name="department">
        <option value="IT">Information Technology</option>
        <option value="IS">Information System</option>
        <option value="CS">Computer Science</option>
    </select>
    <input type="submit" id="loginbtn" name="submit"/>
</form>

信息技术
信息系统
计算机科学

PHP

<?php
require_once("navig.php");
require_once('connect.php');
$dbb = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME)
  or die('Error communicating to MySQL server.');
if(isset($_POST['loginbtn'])){
  if(isset($_POST['department'])){
    $department=$_POST['department'];
    $querye = "INSERT INTO tbl_name(department) ".
                        "VALUES ('$department')";
    $sql=mysqli_query($dbb,$querye);
    mysqli_close($dbb);                     
  }
}
?>

<?php
require_once("navig.php");
require_once('connect.php');
$dbb = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME)
  or die('Error communicating to MySQL server.');
if(isset($_POST['submit'])){
  if(isset($_POST['department'])){
    $department=$_POST['department'];
    $querye = "INSERT INTO tbl_name (department) VALUES ('$department')";
    $sql=mysqli_query($dbb,$querye);
    mysqli_close($dbb);                     
  }
}
?>

---

像这样更改查询

$querye = "INSERT INTO tbl_name(department) VALUES ('$department')";

及

及


更改此查询

$querye = "INSERT INTO tbl_name(department) ".
                        "VALUES ('$department')";

到

也尝试

<input type="submit" id="loginbtn" name="loginbtn" value="Submit" />


让您的查询像-

"INSERT INTO tbl_name(department) VALUES ('$department')";

并设置提交按钮的名称

<input type="submit" id="loginbtn" name="loginbtn" />


试着回应你的$_帖子（“部门”）；如果显示正确的值，则查询有问题

试试这个：

$querye = "INSERT INTO tbl_name(department) VALUES ('$department')";

您的代码中有几个错误。请尝试以下操作：

表格

<form name="stdntdetails" action="study.php" method="post">
    <select name="department">
        <option value="IT">Information Technology</option>
        <option value="IS">Information System</option>
        <option value="CS">Computer Science</option>
    </select>
    <input type="submit" id="loginbtn" />
</form>

<form name="stdntdetails" action="study.php" method="post">
    <select name="department">
        <option value="IT">Information Technology</option>
        <option value="IS">Information System</option>
        <option value="CS">Computer Science</option>
    </select>
    <input type="submit" id="loginbtn" name="submit"/>
</form>


信息技术
信息系统
计算机科学

PHP

<?php
require_once("navig.php");
require_once('connect.php');
$dbb = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME)
  or die('Error communicating to MySQL server.');
if(isset($_POST['loginbtn'])){
  if(isset($_POST['department'])){
    $department=$_POST['department'];
    $querye = "INSERT INTO tbl_name(department) ".
                        "VALUES ('$department')";
    $sql=mysqli_query($dbb,$querye);
    mysqli_close($dbb);                     
  }
}
?>

<?php
require_once("navig.php");
require_once('connect.php');
$dbb = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME)
  or die('Error communicating to MySQL server.');
if(isset($_POST['submit'])){
  if(isset($_POST['department'])){
    $department=$_POST['department'];
    $querye = "INSERT INTO tbl_name (department) VALUES ('$department')";
    $sql=mysqli_query($dbb,$querye);
    mysqli_close($dbb);                     
  }
}
?>

@ABHI_SINGH事实上，把查询保持在一行要容易得多。您没有“提交”按钮的值，正在检查（isset（$\u POST['loginbtn']）。因此，条件失败了。这是您的问题不仅仅是查询尝试设置值以提交按钮，并将该值检查为isset（$_POST['submit button value']）。与您所说的相同，仍然没有值进入数据库。所有连接正确，其他输入ex text工作正常，只需一次尝试，$querye=“插入tbl_名称（部门）值（“$department”。”）；尝试此查询您确定在
$\u POST['department']
中获得了值。试试echo$_POST['department']将此问题编辑到您编辑的代码中。@ABHI_SINGH，在
下面如果（设置（$\u POST['loginbtn']）{
，放置
echo“Success”
。显示结果！。编辑代码。在
$department=$\u POST['department]下面再次尝试成功，但仍然没有获得存储的价值
put
echo$department；
您得到了什么？@ABHI\u SINGHwhen value是submit，echo时没有得到任何东西，而value是login，btn也没有得到任何东西。请帮助为什么要使用2倍id元素
id=“loginbtn”
。删除它。而不是将其放入
name=“loginbtn”
。如果您没有获得
$department
的值，那么如何将其存储在数据库中？