-1
我有一個名爲FACULTY
的表,其中有FacultyID
,FName
和LName
。它預先填充了信息。我可以使用動態下拉式來將這些值顯示在頁面上。如何使用php發送ID到數據庫的動態下拉
我的問題是我也有一個正在連接到每個教師的出版物表。我希望能夠選擇多個教師並將這些ID(單個行)發送到教師發佈表。我希望輸入的發佈信息進入發佈表,然後自動生成的ID也進入教師發佈表。
表名:FACULTY
,PUBLICATION
,FACULTYPUBLICATIONS
這裏是我到目前爲止的代碼:
<?php
include_once 'dbc.php';
function connect() {
mysql_connect(DB_HOST, DB_USER, DB_PASS) or die('Could not connect to database'.mysql_error());
mysql_select_db(DB_NAME);
}
function close() {
mysql_close();
}
function query() {
$myData = mysql_query("SELECT * FROM FACULTY");
while($record = mysql_fetch_array($myData)) {
echo '<option value="'. $record['FName'] .$record['LName']. '">' . $record['FName'] .' '. $record['LName'] . ' </option>';
}
}
<?php
include_once 'pullDataR2.php';
connect();
?>
<!DOCTYPE html>
<head>
<link href="styles.css" rel="stylesheet">
<h1> help </h1>
</head>
<body>
<div class="StyleDiv" >
<!-- coding for journal -->
<form id="form1" name="form1" method="post" action="RR2.php">
<label for="FName">Faculty Name</label>
<select multiple="multiple" name="select" id="Faculty">
<?php query() ?>
</select>
<?php close() ?>
<br class="clear" />
<input type="hidden" name="JournalID" id="JournalID" class="textbox" />
<br class="clear" />
<label for="JournalName">Journal Name</label>
<input type="text" name="JournalName" id="JournalName" />
<br class="clear" />
<label for="Rating">Journal Rating</label>
<select name="Rating" id="Rating">
<option value="A+">A+</option>
<option value="A">A</option>
<option value="A-">A-</option>
<option value="B+">B+</option>
<option value="B">B</option>
<option value="B-">B-</option>
<option value="C+">C+</option>
<option value="C">C</option>
<option value="C-">C-</option>
<option value="D+">D+</option>
<option value="D">D</option>
<option value="D-">D-</option>
<option value="F">F</option>
</select>
<br class="clear" />
<!-- coding for publication -->
<input type="hidden" name="PubID" id="PubID" />
<br class="clear" />
<label for="Title">Publication Title</label>
<input type="text" name="Title" id="Title" />
<br class="clear" />
<label for="Year">Year</label>
<input type="text" name="Year" id="Year" />
<br class="clear" />
<label for="Volume">Volume</label>
<input type="text" name="Volume" id="Volume" />
<br class="clear" />
<label for="Issue">Issue</label>
<input type="text" name="Issue" id="Issue" />
<br class="clear" />
<label for="Comments">Comments</label>
<textarea name="Comments" id="Comments" cols="45" rows="5"></textarea>
<br class="clear" />
<input type="submit" name="Submit" id="Submit" value="Submit" />
<br class="clear" /></br></br>
</div>
</form>
<?php
//Post Parameters
$JournalName = $_POST['JournalName'];
$Rating = $_POST['Rating'];
$Year = $_POST['Year'];
$Comments = $_POST['Comments'];
$Volume = $_POST['Volume'];
$Issue = $_POST['Issue'];
$Title = $_POST['Title'];
//create connection
$conn = new mysqli('localhost','root','isasurvey','isasurvey');
if($conn->connect_errno) {
echo 'failure</br>';
}
//Query
//INSERT
$stmt = $conn->prepare (" INSERT INTO JOURNAL (JournalName, Rating) VALUES ('$JournalName', '$Rating')");
$stmt->bind_param("sssssss", $JournalName, $Rating);
$stmt->execute();
//INSERT
$stmt = $conn->prepare(" INSERT INTO PUBLICATION (Year, Comments, Volume, Issue, Title) VALUES ('$Year', '$Comments', '$Volume', '$Issue', '$Title')");
$stmt->bind_param("sssssss", $Year, $Comments, $Volume, $Issue, $Title);
$stmt->execute();
?>
</body>
</html>
你有什麼問題?你試過什麼了? – Barett 2015-04-02 20:31:02
我無法獲得FacultyID以進入發佈表。當我點擊提交時 - 它進入出版和期刊很好。現在,我需要將自動生成的發佈ID放入FacultyPublications並將其鏈接到該預填表中的FacultyID(特別是如果您選擇多個教師)。我試圖插入一個循環,但只是使頁面消失。我想知道我是否可以使用uuid,但我仍然不知道如何讓多個教師工作。 因此,如果我有出版物1 - 它由2個教職員編寫,需要2行。 – reedma1 2015-04-02 22:18:01
我可以使用請求來實現這個功能嗎? – reedma1 2015-04-05 14:43:26