HTML下拉菜单不会生成所选的MySQL表
HTML下拉菜单不会生成所选的MySQL表
提问于 2018-06-05 08:03:36
我正在尝试使用此表单来显示数据库中的表。我不确定我做错了什么。我交换了"form“和"select”标签的位置，并假定这就是修复方法。但每次从下拉列表中选择任何内容时，页面都会简单地重新加载我创建的普通数据库。正如您所看到的，我在php中填充我的表，我唯一的问题是当我尝试将该信息放入HTML表单中时，它不起作用。我可以给你发送我的意思的例子，这会有所帮助，但我假设你明白我想要做的事情。
<!-- Use JavaScript to automatically submit the selection -->
<select name="lstDisplay" onchange="this.form.submit()">
<option value="null">Select an item</option>
<option value="concert">Concert</option>
<option value="attendee">Atendee</option>
<option value="venue">Venue</option>
</select>
<!-- set up alternative button in case JavaScript is not active -->
<noscript>
<input type="submit" name="btnSubmit" value="View the list" />
<br /><br />
</noscript>
<!-- Use a hidden field to tell server if return visitor -->
<input type="hidden" name="hidIsReturning" value="true" />
</form>
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
// Start with a new database to start primary keys at 1
$sql = "DROP DATABASE " . DATABASE_NAME;
runQuery($sql, "DROP " . DATABASE_NAME, true);
// Create database if it doesn't exist
$sql = "CREATE DATABASE IF NOT EXISTS " . DATABASE_NAME;
//if ($conn->query($sql) === TRUE) {
// echo "The database " . DATABASE_NAME . " exists or was created succesffuly!<br/>";
//}
//else {
// echo "Error creating database " . DATABASE_NAME . ": " . $conn->error;
// echo "<br/>";
//}
runQuery($sql, "Creating " . DATABASE_NAME, false);
// Select the database
$conn->select_db(DATABASE_NAME);
/*
--------------------------
* Create the tables
--------------------------
*/
// Create Table: volunteer
/*
--------------------------
* Create the tables
--------------------------
*/
// Create Table: attendee
$sql = "CREATE TABLE IF NOT EXISTS attendee (
attendee_id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
fName VARCHAR(20),
lName VARCHAR(20),
phone VARCHAR(15),
email VARCHAR(50)
)";
runQuery($sql, "Table:attendee", false);
// Create Table: concert
$sql = "CREATE TABLE IF NOT EXISTS concert (
concert_id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
concert VARCHAR(20) NOT NULL
)";
runQuery($sql, "Table:concert", false);
// Create Table: attendee_concert
$sql = "CREATE TABLE IF NOT EXISTS attendee_concert (
attendee_concert_id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
attendee_id INT(6) UNSIGNED,
concert_id INT(6) UNSIGNED,
paid TINYINT(1)
)";
runQuery($sql, "Table:attendee_concert", false);
// Create Table: venue
$sql = "CREATE TABLE IF NOT EXISTS venue (
venue_id INT(4) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
venueName VARCHAR(25)
)";
runQuery($sql, "Table:venue", false);
/*
--------------------------------------------
* Populate Tables Using Sample Data attendee
* This data will later be collected using a form.
--------------------------------------------
*/
// Populate table: attendee
$attendeeArray = array(
array("Rob", "Nelson", "651-333-3030", "Rob@gmail.com"),
array("Matt", "Doe", "888-867-5309", "Matt@gmail.com"),
array("Tom", "Reynolds", "651-303-9090", "Tom@gmail.com"),
array("Jane", "Doe", "651-678-8901", "Jane@gmail.com"),
array("Emily", "Nelson", "612-234-5678", "Emily@gmail.com"),
array("Timmy", "Turner", "987-098-0987", "Timmy@gmail.com")
);
foreach($attendeeArray as $attendee) {
echo $attendee[0] . " " . $attendee[1] . "<br/>";
$sql = "INSERT INTO attendee (fName, lName, phone, email) "
. "VALUES ('" . $attendee[0] . "', '"
. $attendee[1] . "', '"
. $attendee[2] . "', '"
. $attendee[3] . "')";
runQuery($sql, "Record inserted for: " . $attendee[0], false);
}
// Populate Table: concert
$concertArray = array("Brand New", "Thrice", "Daft Punk", "Kanye West",);
foreach($concertArray as $concert) {
$sql = "INSERT INTO concert (concert) " . "VALUES ('" . $concert . "')";
runQuery($sql, "New record insert $concert[0]", false);
}
// Populate Table: attendee_concert
$attendee_concertArray = array(
array(1,1,1),
array(2,2,1),
array(3,3,1),
array(4,3,1),
array(5,3,1),
array(6,4,1)
);
foreach ($attendee_concertArray as $attendee_concert) {
$sql = "INSERT INTO attendee_concert (attendee_id, concert_id, paid) "
. "VALUES ('" . $attendee_concert[0] . "', '"
. $attendee_concert[1] . "', '"
. $attendee_concert[2] . "')";
runQuery($sql, "New record insert $attendee_concert[0]", false);
}
// Populate Table: venue
$venueArray = array("The Myth", "Target Field", "The Cabooze", "Blue Door Pub");
foreach ($venueArray as $venue) {
$sql = "INSERT INTO venue (venueName) "
. "VALUES ('" . $venue . "')";
runQuery($sql, "New record insert $venue[0]", true);
}
$sql = "SELECT * FROM attendee";
$result = $conn->query($sql);
displayResult($result, $sql);
$conn->close();
function runQuery($sql, $msg, $echoSuccess) {
global $conn;
// run the query
if ($conn->query($sql) === TRUE) {
if($echoSuccess) {
echo $msg . " successful.<br/>";
}
} else {
echo "<strong>Error when: " . $msg . "</strong> using SQL: " . $sql . "<br/>" . $conn->error;
}
} // end of runQuery()
function displayResult($result, $sql) {
if ($result->num_rows > 0) {
echo "<table border='1'>\n";
// print headings (field names)
$heading = $result->fetch_assoc( );
echo "<tr>\n";
// Print field names as table headings
foreach($heading as $key=>$value){
echo "<th>" . $key . "</th>\n";
}
echo "</tr>";
// Print the values for the first row
echo "<tr>";
foreach($heading as $key=>$value){
echo "<td>" . $value . "</td>\n";
}
// Output each record
while($row = $result->fetch_assoc()) {
//print_r($row);
//echo "<br />";
echo "<tr>\n";
// print data
foreach($row as $key=>$value) {
echo "<td>" . $value . "</td>\n";
}
echo "</tr>\n";
}
echo "</table>\n";
// No results
} else {
echo "<strong>zero results using SQL: </strong>" . $sql;
}
} // end of displayResult( )
?>回答 1
Stack Overflow用户
发布于 2018-06-05 08:08:16
您的<select>有onchange="this.form.submit()"，它告诉JavaScript在您做出选择时提交表单。只需删除它，以防止您的表单自动提交。此外，您还希望您的提交按钮位于<noscript>之外(这实际上根本不是必需的)。
最终，您希望您的<form>看起来如下所示：
<form>
<select name="lstDisplay">
<option value="null">Select an item</option>
<option value="concert">Concert</option>
<option value="attendee">Atendee</option>
<option value="venue">Venue</option>
</select>
<input type="submit" name="btnSubmit" value="View the list" />
<!-- Use a hidden field to tell server if return visitor -->
<input type="hidden" name="hidIsReturning" value="true" />
</form>
假设您希望在没有任何类型的用户交互的情况下自动将数据发布到您的服务器，那么您应该使用而不是表单。
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/50690307
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