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我一直試圖解決這個問題幾個小時,但仍然沒有進展。想要將我的數據庫結果回覆到在我的html中創建的表格中,但它只是彈出窗口一側。請建議我可以做的事情。謝謝。 PHPPHP的回聲表不工作
if (isset($_POST['search'])) {
include 'dbh.php';
$uid =$_POST['name1'];
$nric = $_POST['nric1'];
$number = $_POST['number1'];
$sql = "SELECT * FROM user WHERE name = '$uid' OR nric = '$nric' OR contact = '$number' ";
$result = $conn->query($sql);
if (!$row = mysqli_fetch_assoc($result))
{
echo "No result is found! ";
}
else
{
echo "result found";
$_SESSION['id'] = $row['id'];
echo "<tr>";
echo "<td>".$row['no']."</td>";
echo "<td>".$row['name']."</td>";
echo "<td>".$row['surname']."</td>";
echo "<td>".$row['nric']."</td>";
echo "<td>".$row['contact']."</td>";
echo "<td>".$row['gender']."</td>";
echo "<td>".$row['email']."</td>";
echo "<td>".$row['address']."</td>";
echo "<td>".$row['date']."</td>";
echo "</tr>";
}
}
在我的形式方法。
<!-- Search function aboutus about us website -->
<form method="post" action="" >
<div class="panel" id="aboutus">
<h1>About Us</h1>
<!-- end of header -->
<input type="text" name="nric1"
placeholder="NRIC"><br> <input type="text" name="name1"
placeholder="Name"><br>
<input type="text" name="number1" placeholder="Details Number"><br>
<table width = "300" border = "1" cellpadding="1" cellspacing="1">
<tr>
<th>No</th>
<th>Name</th>
<th>Surname</th>
<th>Nric</th>
<th>Contact</th>
<th>Gender</th>
<th>Email</th>
<th>Address</th>
<th>Data</th>
<tr>
</table>
<input type="submit" name="search" value="Search">
</form>