【发布时间】：2018-06-05 00:03:36
【问题描述】：
我正在尝试使用此表单来显示我的数据库中的表格。我不确定我做错了什么。假设这是修复，我交换了“表单”和“选择”标签的位置。但是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的正常数据库创建。 如您所见，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 表单时它不起作用。我可以向您发送我的意思的示例，这将有所帮助，但我假设您得到了我想要做的事情。
<!-- 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( )
?>
【问题讨论】：
-
为什么选择元素上有 onchange="this.form.submit()"。您还有一个提交表单的输入按钮。表单的意图不是让用户选择一个值然后点击提交按钮吗？
-
这和phpMyAdmin有什么关系？
-
（另外，我希望您不会在每个页面浏览量上删除/创建 MySQL 表...）
-
我只是想使用 HTML 表单来显示我在 phpmyadmin 中创建的数据库中的表格。出于明显的原因，我没有把数据库名称放在上面。
-
这段代码与 phpMyAdmin 无关。 phpMyAdmin 只是一个编辑 MySQL 数据库的工具。我将您的标签/标题改为
mysql。
标签： php mysql select html.dropdownlistfor database-table