下面給出的PHP函數顯示最新插入一個頁面加載,但是我希望每次插入行時刷新數據。我聽說過阿賈克斯,但我並不是很熟悉它。任何人都可以給我一些指導嗎?僅當發生插入時才使用PHP顯示來自MYSQL的記錄
<!DOCTYPE html>
<html>
<head>
<link rel="stylesheet" href="OPlayout.css"/>
<title>Operator View</title>
</head>
<body>
<div id="whole">
<section id="mid_section"></br>
<h2>Requests</h2></br>
<?php include 'functions.php';
currentRequests();
?>
</section>
</div>
</body>
</html>
PHP函數:
function currentRequests(){
//Include database connection file
include 'databaseConnect.php';
//Call global function to connect to db
$connection = connectToDatabase();
$queryToRun = "SELECT * FROM table_user_request
WHERE 1=1
AND date(entry_date) = date(now())
ORDER BY entry_date DESC";
$result = mysqli_query($connection, $queryToRun);
//Create a table to store the results
echo "<table border = '1' style='max-width:1180px;font-size: 14px;' >
<tr>
<td style='width:120px '>Name</td>
<td style='width:120px '>Location</td>
<td style='width: 20px '># of Riders</td>
<td style='width:120px '>Destination</td>
<td style='width:150px '>Email</td>
<td style='width:100px '>Telephone</td>
<td style='width:50px '>Date</td>
<td style='width:50px '>Status</td>
<td style='width: 60px '>Vehicle</td>
<td style='width:50px '>Cancellation Reason</td>
<td style='width: 60px '>Priority</td>
</tr>";
//Print records until done
while ($row = mysqli_fetch_array($result, MYSQLI_NUM)) {
//echo("<table border = '1' style='width:1200px'>");
//Output results found in the database
echo("<tr>
<td>" . $row['1']
. "</td> <td>" . $row['3']
. "</td> <td>" . $row['5']
. "</td> <td>" . $row['4']
. "</td> <td>" . $row['8']
. "</td> <td>" . $row['9']
. "</td> <td>" . $row['10']
. "</td> <td>" . $row['11']
. "</td> <td>" . $row['12']
. "</td> <td>" . $row['13']
. "</td> <td>" . $row['14'] . "</td>
</tr>");
}
echo("</table>");
}
如何插入新記錄?我沒有看到HTML/JS代碼。它是否在同一頁面上完成? – stackErr
它是在不同的頁面上完成的,這是一個呼叫用戶界面。一旦通過用戶界面插入記錄,我希望能夠在操作員的一端顯示它,而不必刷新整個頁面。 – jorgeAChacon
如果數據庫中存在更多或更少的行,並且相應地更新頁面,則可以每隔幾分鐘比較一次。 – stackErr