在 PHP MYSQL 中獲取一個按鈕來顯示查詢結果

Question

我試圖運行一個查詢，當我單擊按鈕時，該查詢將顯示在我的瀏覽器中，但結果只是從頭開始顯示。 我將按鈕設置為發布，並將我的 php 文件設置為操作，但它似乎只是從一開始就運行代碼。 這是我所擁有的：

<?php
$host = "localhost";
$db = "cis475";
$user = "root";
$pw = "";


$conn = new mysqli ($host, $user, $pw, $db);
if($conn->connect_error) die($conn->connect_error);

$readAllQuery = "SELECT * FROM enrollment JOIN course ON enrollment.CourseID 
= course.CourseID JOIN student ON enrollment.StudentID = student.StudentID";

    $result = $conn->query($readAllQuery);
    if (!$result) die($conn->error);

    echo "<table border='1'>";
    echo "<tr><td>EnrollmentID</td><td>Grade</td><td>EnrollmentSemester</td> 
   <td>CourseID</td><td>StudentID</td><td>Title</td><td>Credits</td> 
   <td>LastName</td><td>FirstMidName</td></tr>";
    while ($row = mysqli_fetch_assoc($result)){
        echo "<tr><td>{$row['EnrollmentID']}</td><td>{$row['Grade']}</td> 
   <td>{$row['EnrollmentSemester']}</td><td>{$row['CourseID']}</td><td> 
   {$row['StudentID']}</td><td>{$row['Title']}</td><td>{$row['Credits']} 
   </td> 
   <td>{$row['LastName']}</td><td>{$row['FirstMidName']}</td></tr>";

    }
echo "</table>";

?>

<form method='post' action='readAll.php'>

<input type='submit' name='submit' value='Show All Enrollments'>

</form>

Answer 1

您需要查看是否存在“某物”（在這種情況下，您所擁有的只是提交），然后進行顯示....

<?php
if($_POST['submit']){
    $host = "localhost";
    $db = "cis475";
    $user = "root";
    $pw = "";

    $conn = new mysqli ($host, $user, $pw, $db);
    if($conn->connect_error) die($conn->connect_error);

    $readAllQuery = "SELECT * FROM enrollment JOIN course ON enrollment.CourseID = course.CourseID JOIN student ON enrollment.StudentID = student.StudentID";

    $result = $conn->query($readAllQuery);
    if (!$result) die($conn->error);

    echo "<table border='1'>";
    echo "<tr><td>EnrollmentID</td><td>Grade</td><td>EnrollmentSemester</td> 
    <td>CourseID</td><td>StudentID</td><td>Title</td><td>Credits</td> 
    <td>LastName</td><td>FirstMidName</td></tr>";
    while ($row = mysqli_fetch_assoc($result)){
        echo "<tr><td>{$row['EnrollmentID']}</td><td>{$row['Grade']}</td> 
        <td>{$row['EnrollmentSemester']}</td><td>{$row['CourseID']}</td><td> 
        {$row['StudentID']}</td><td>{$row['Title']}</td><td>{$row['Credits']} 
        </td><td>{$row['LastName']}</td><td>{$row['FirstMidName']}</td></tr>";
    }
    echo "</table>";
}
?>

<form method='post' action='readAll.php'>

<input type='submit' name='submit' value='Show All Enrollments'>

</form>

Answer 2

程序文件中的任何代碼都將在執行時運行，這對所有語言都是如此，除非它包含在if或任何其他條件語句中； 在這種情況下，包含在條件語句中的代碼只有在條件為true才會執行。

因此，為了解決您的問題，您需要一個條件語句來檢查表單是否已提交，是否已提交，然后執行以下代碼-

if(isset($_POST['submit'])){
    $host = "localhost";
    $db = "cis475";
    $user = "root";
    $pw = "";

    $conn = new mysqli ($host, $user, $pw, $db);
    if($conn->connect_error) die($conn->connect_error);

    $readAllQuery = "SELECT * FROM enrollment JOIN course ON enrollment.CourseID = course.CourseID JOIN student ON enrollment.StudentID = student.StudentID";

    $result = $conn->query($readAllQuery);
    if (!$result) die($conn->error);

    echo "<table border='1'>";
    echo "<tr><td>EnrollmentID</td><td>Grade</td><td>EnrollmentSemester</td> 
    <td>CourseID</td><td>StudentID</td><td>Title</td><td>Credits</td> 
    <td>LastName</td><td>FirstMidName</td></tr>";
    while ($row = mysqli_fetch_assoc($result)){
        echo "<tr><td>{$row['EnrollmentID']}</td><td>{$row['Grade']}</td> 
        <td>{$row['EnrollmentSemester']}</td><td>{$row['CourseID']}</td><td> 
        {$row['StudentID']}</td><td>{$row['Title']}</td><td>{$row['Credits']} 
        </td><td>{$row['LastName']}</td><td>{$row['FirstMidName']}</td></tr>";
    }
    echo "</table>";
}

在這個片段中，我們使用if語句來檢查變量$_POST['submit']存在。
這里submit with 是表單提交按鈕的名稱
注意： $_POST _ $_POST變量僅在表單已提交時才存在，如果表單未提交，則調用isset()將返回false ，因此您在{ }的代碼不會在第一次加載時執行，但會在提交時運行。