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我想嘗試在我的網頁上添加兩個過濾器。使用單選按鈕過濾數據
- 一個日期過濾器和另一個過濾器的顏色。
- 如果未設置過濾器，則應顯示所有記錄。
我想使用if/else語句以及名稱/值對。
任何人都可以看到我做錯了什麼？我嘗試過幾件事情，而這正是我所結束的。我也讀過很多頁面，並嘗試使用這些例子，但似乎沒有任何工作適合我。 Stackoverflow linkStackoverflow link
HTML/PHP
<?php
//connect to the database
$dbc = mysqli_connect('host', 'user', 'password', 'cars') or die('Error connecting to MySQL Server.');
//If RadioButton Clicked Sort the Database by dateadded Asc/Desc
if(isset($_POST['dateorder'])){
if($_POST['dateorder'] == 'dateasc'){
//Run query for dateasc
$query = "SELECT * FROM cardetails ORDER BY caradded asc";
}elseif($_POST['dateorder'] == 'datedesc'){
//Run query for datedesc
$query = "SELECT * FROM cardetails ORDER BY caradded desc";
}
}else{
$query = "SELECT * FROM cardetails ORDER BY id asc";
}
//If RadioButton Clicked Sort the Database by Color Red, Green, Blue
if(isset($_POST['color'])){
if($_POST['color'] == 'red'){
//Run query for red color
$query = "SELECT * FROM cardetails WHERE color = 'red'";
}elseif($_POST['color'] == 'green'){
//Run query for green color
$query = "SELECT * FROM cardetails WHERE color = 'green'";
}elseif($_POST['color'] == 'blue'){
//Run query for blue color
$query = "SELECT * FROM cardetails WHERE color = 'blue'";
}
}else{
$query = "SELECT * FROM cardetails ORDER BY id asc";
}
$result = mysqli_query($dbc, $query) or die('Error Refreshing the page: ' . mysqli_error($dbc));
//Retrieve the practice tasks from the database table
$result = mysqli_query($dbc, $query) or die('Error querying database.');
//start pagination
//Get the total count of rows.
$sql = "SELECT COUNT(id) FROM cardetails";
$query = mysqli_query($dbc, $sql);
$row = mysqli_fetch_row($query);
//Here we have the total row count
$rows = $row[0];
//Number of results to show per page
$page_rows = 5;
//This tells us the page number of our last page
$last = ceil($rows/$page_rows);
//This makes sure last cannot be less than 1
if($last < 1){
$last = 1;
}
//establish the $pagenum variable
$pagenum = 1;
//Get pagenum from URL vars if it is present, else it is = 1
if(isset($_GET['pn'])){
$pagenum = preg_replace('#[^0-9]#', '', $_GET['pn']);
}
//This makes sure the page number isn't below 1 or more than our $last page
if($pagenum < 1){
$pagenum = 1;
} else if($pagenum > $last) {
$pagenum = $last;
}
//This sets the range of rows to query for the chosen $pagenum
$limit = 'LIMIT ' . ($pagenum - 1) * $page_rows .',' .$page_rows;
//This is your query again, it is for grabbing just one page worth of rows by applying the limit variable
$sql = "SELECT id, caradded, make, color FROM cardetails ORDER BY id DESC $limit";
$query = mysqli_query($dbc, $sql);
//$paginationCrls variable
$paginationCtrls = ' ';
//if there is more than 1 page worth of results
if($last != 1){
/*First we check if we are on page one. If we are then we don't need a link to
the previous page or the first page so we do nothing. If we aren't then we generate
links to the first page and to the previous page.*/
if($pagenum > 1){
$previous = $pagenum - 1;
$paginationCtrls .= '<a href="' . $_SERVER['PHP_SELF']. '?pn=' .$previous. '">« Previous</a> ';
//Render clickable number links that should appear on the left of the target page number
for($i = $pagenum-4; $i < $pagenum; $i++){
if($i > 0) {
$paginationCtrls .= '<a href="' .$_SERVER['PHP_SELF']. '?pn=' . $i .'">'.$i.'</a> ';
}
}
}
//Render the target page number, but without it being a link
$paginationCtrls .= ''.$pagenum.' ';
//Render clickable number links that should appear on the right of the target page number
for($i = $pagenum+1; $i <= $last; $i++){
$paginationCtrls .= '<a href="'.$_SERVER['PHP_SELF'].'?pn='.$i.'">'.$i.'</a> ';
if($i >= $pagenum+4){
break;
}
}
//This adds the same as above only checking if if we are on the last page and then generating Next
if($pagenum != $last){
$next = $pagenum + 1;
$paginationCtrls .= ' <a href="'.$_SERVER['PHP_SELF'].'?pn='.$next.'">Next »</a> ';
}
}
//Finish Pagination
?>
<!DOCTYPE html>
<html>
<head>
<title>Basic Page Layout</title>
<link rel="stylesheet" type="text/css" href="basicstyle.css" />
</head>
<body>
<div id="wrapper">
<div id="headerwrap">
<div id="header">
<p>Search/Filter Practice</p>
</div>
</div>
<div id="leftcolumnwrap">
<div id="leftcolumn">
<h2>Trial Filters</h2>
<form method="post" action="<?php echo htmlentities($_SERVER['PHP_SELF']); ?>">
<p>Filter by Date:</p>
<input type="radio" name="dateorder" value="dateasc"><label for="dateasc">A - Z</label><br>
<input type="radio" name="dateorder" value="datedesc"><label for="datedesc">Z - A</label><br>
<br><hr>
<p>Filter by Colour:</p>
<input type="radio" name="color" value="red"><label for="red">Red</label><br>
<input type="radio" name="color" value="green"><label for="green">Green</label><br>
<input type="radio" name="color" value="blue"><label for="blue">Blue</label>
<br><br>
<input name="submit" type="submit">
<br><br>
</div>
</div>
<div id="contentwrap">
<div id="content">
<p style="text-align:center; font-weight:bold;">My Database Records</p>
<?php
echo "<table>";
echo "<tr>";
echo "<th>Car Added On</th>";
echo "<th>Make</th>";
echo "<th>Color</th>";
echo "</tr>";
while($row = mysqli_fetch_array($query)){
echo "<tr>";
echo "<td>". $row['caradded'] ."</td>";
echo "<td>". $row['make'] ."</td>";
echo "<td>". $row['color'] ."</td>";
echo "</tr>";
}
print '</table><br><br>';
//close connection to database
mysqli_close($dbc);
?>
<p id="pagination_controls"><?php echo $paginationCtrls;?><br><br>
</div>
</div>
<div id="footerwrap">
<div id="footer">
<p>July/August 2016</p>
</div>
</div>
</div>
</body>
</html>
MySQL的
CREATE DATABASE cars;
USE cars;
CREATE TABLE cardetails(
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
caradded datetime NOT NULL DEFAULT NOW(),
make VARCHAR(50) NOT NULL,
color VARCHAR(50) NOT NULL);
DESCRIBE cardetails;
INSERT INTO cardetails(caradded, make, color) VALUES
(NOW(), "Toyota", "Red"),
(NOW(), "VW", "Green"),
(NOW(), "Nissan", "Blue"),
(NOW(), "Toyota", "Green"),
(NOW(), "Toyota", "Green"),
(NOW(), "VW", "Blue"),
(NOW(), "VW", "Green"),
(NOW(), "Nissan", "Blue"),
(NOW(), "Nissan", "Green");
SELECT * FROM cardetails;
什麼是您的具體問題？什麼不行？ –
嗨盧卡，當我點擊無論是asc還是desc單選按鈕，沒有任何反應，我需要更多的代碼，幫助找出答案。 –
@LucaJung更新了我的例子，如果你有任何想法。 –