如何將產生的按鈕的值從一個 PHP 檔案傳遞到另一個 PHP 檔案?
P粉155832941
P粉155832941 2023-09-14 21:00:35
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我想將 PHP 產生的按鈕的 id 值從theory.php檔案傳遞到theory1.php檔案。 程式碼如下:

//theory.php file require('components/db.php'); $query = "SELECT * FROM `courses`"; $result = mysqli_query($connect, $query) or die("Error:" . mysqli_error($connect));; $numrows = mysqli_num_rows($result); for ($i = 0; $i < $numrows; $i++) { $query = "SELECT * FROM `courses` WHERE courseID = '$i'"; $result = mysqli_query($connect, $query) or die("Error:" . mysqli_error($connect));; $rowQuery = mysqli_fetch_assoc($result); $_SESSION['course_ID'] = $i; echo ' 
Course 1

'; echo $rowQuery['courseName']; echo '

'; echo $rowQuery['courseTextOne']; echo '

Proceed
'; //a - is a button which needs to have an ID to pass to theory1.php }

該程式碼產生帶有按鈕的卡片。我希望每個按鈕都儲存 MySQL 資料庫中課程的相應 ID。該 ID 需要根據單擊的按鈕(卡)傳遞到另一個頁面,以便將來可以從資料庫中檢索正確的資料。

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解決方案

理論.php:

#
require('components/db.php'); $query = "SELECT * FROM `courses`"; $result = mysqli_query($connect, $query) or die("Error:" . mysqli_error($connect));; $numrows = mysqli_num_rows($result); for ($i = 0; $i < $numrows; $i++) { $query = "SELECT * FROM `courses` WHERE courseID = '$i'"; $result = mysqli_query($connect, $query) or die("Error:" . mysqli_error($connect));; $rowQuery = mysqli_fetch_assoc($result); $_SESSION['course_ID'] = $i; echo ' 
Course 1

'; echo $rowQuery['courseName']; echo '

'; echo $rowQuery['courseTextOne']; echo '

Перейти
'; //a - is a button which needs to have an ID to pass to theory1.php }

theory1.php:

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