我有一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，以便我可以获得所选会议的变量。我是数据库的新手，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等于所选选项的变量。表单已提交，但我没有任何价值。我查遍了整个网络，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 connectionif (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>$row[name]</option>";}echo "</select>";//Close phpMyAdminmysqli_close($con);?><script> $( "#searchddl" ).chosen()</script><?phpecho $db_table;?></body></html>
1 回答
慕运维8079593
TA贡献1876条经验 获得超5个赞
你没有value在你<option>的属性中给出属性，这就是$db_table当表单被 sumbitted 时没有传入任何内容的原因。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
}
?>
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