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我有这个显示不同会议名称的下拉列表。我可以选择某个会议，但是当我选择会议时，我希望能够单击提交按钮，以便获得所选会议的变量。

我是数据库新手，但我尝试添加一个表单，但我似乎无法让它在 PHP 代码中工作。数据库连接并显示所有会议都很好，我只是不知道如何获得一个等于所选选项的变量。表单被提交，但我没有得到任何价值。我浏览了整个网络，但一无所获。

.error {
  color: #FF0000;
}
<!DOCTYPE HTML>  
<html>
<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.4.1/jquery.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/chosen/1.8.7/chosen.jquery.min.js"></script>
<link rel="stylesheet" href="https://cdnjs.cloudflare.com/ajax/libs/chosen/1.8.7/chosen.min.css" />
</head>
<body>  
<form action="" name="selection" method="post">
<select project="ConferenceList" id="ConferenceList" name="ConferenceList">

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

<?php
//Declare variables
$db_host = "";
$db_username = "";
$db_pass = "";
$db_name = "";
//$db_table = "";
//Connect to phpMyAdmin
$con=mysqli_connect("$db_host","$db_username","$db_pass","$db_name");

// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

mysqli_select_db($con,"$db_name") or die ("No database");

$result=mysqli_query($con,"select * From conferenceList");

echo "<select id='searchddl'>";
echo "<option> -- Search Conference Name -- </option>";
while($row=mysqli_fetch_array($result))
{
    echo "<option>$row[name]</option>";
}
echo "</select>";

//Close phpMyAdmin
mysqli_close($con);
?>


<script>
    $( "#searchddl" ).chosen()
</script>


<?php
echo $db_table;
?>

</body>
</html>

4

1 回答 1

0

您没有value在您的中给出属性，这就是当表单被提交时<option>没有传递任何内容的原因。$db_table而是像下面这样：

  <?php
    //Declare variables
    $db_host = "";
    $db_username = "";
    $db_pass = "";
    $db_name = "";
    //$db_table = "";
    //Connect to phpMyAdmin
    $con=mysqli_connect("$db_host","$db_username","$db_pass","$db_name");

    // Check connection
    if (mysqli_connect_errno())
      {
      echo "Failed to connect to MySQL: " . mysqli_connect_error();
      }

    mysqli_select_db($con,"$db_name") or die ("No database");


    ?>

    <form action="" name="selection" method="post">
    <select project="ConferenceList" id="ConferenceList" name="ConferenceList">
    <?php 
    $result=mysqli_query($con,"select * From conferenceList");
    echo "<option> -- Search Conference Name -- </option>";
    while($row=mysqli_fetch_array($result))
    {
      //when form will get submitted whatever will be in value get passed
        echo "<option value='.$row[name].'>$row[name]</option>";
    }
    echo "</select>";

    //Close phpMyAdmin
    mysqli_close($con); ?>

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

     <?php
        $db_table = "";
     //checking if form is submit 
        if (isset($_POST["submit"])) {
            $db_table = $_POST["ConferenceList"];//will give you value of option selected
         echo $db_table;//printing value 
        }
      ?>
于 2019-05-23T05:47:40.273 回答