我有一個顯示不同會議名稱的下拉列表。我可以選擇某個會議，但是當我選擇會議時，我希望能夠單擊提交按鈕，以便我可以獲得所選會議的變量。我是數據庫的新手，但我嘗試添加一個表單，但我似乎無法讓它在 PHP 代碼中工作。數據庫連接并顯示所有會議都很好，我只是不知道如何獲得等于所選選項的變量。表單已提交，但我沒有任何價值。我查遍了整個網絡，但一無所獲。.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 $db_table;?></body></html>
1 回答
慕運維8079593
TA貢獻1876條經驗 獲得超5個贊
你沒有value在你<option>的屬性中給出屬性，這就是$db_table當表單被 sumbitted 時沒有傳入任何內容的原因。而是像下面這樣：
<?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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