Question

我一直试图解决这个问题几个小时但仍然没有进展。想要将我的数据库结果回显到我的html中创建的表格，但它只是弹出窗口的一侧。请建议我能做的事情。谢谢。 PHP

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>";
    }

}

在我的表单方法中。

      <!-- 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>

Answer 1

当然它确实如此，你需要在HTML内部的适当位置回显它。假设你还有HTML inisde相同的php页面

<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>
<?php  ...else{
            $_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>";
?>


     </table>
     <input type="submit" name="search" value="Search">
    </form>

php echo表无法正常工作

1 个答案: