Question

我正在尝试打印一张包含学生推荐书的表格，我想打印出ID，标题，文字和学生ID，但是当点击＆＃34; search_all＆＃34;按钮，我被告知它无法获取mysqli，我不知道该怎么做。

<?php ("session.php"); ?>
<?php require_once("connect.php"); ?>
<?php include("header.php"); ?>             

<html>
    <body>

    <!-- Page Content -->
                <h1 class="page-header">Learning Journals
                    <small>- Admin Search</small>
                </h1>

                <h3>Search Learning Journals</h3>

                <form name="membership_form" action= "<?php echo htmlspecialchars($_SERVER["PHP_SELF"]);?>" id="contactForm"  method="post">

                    <label>Search Journals ID:</label>
                    <br>
                        <input type="text" name="journal_search">
                    <br>

                    <button type="submit" name ="search_all" class="btn btn-primary">All Journals</button>

                </form>    
<?php 

    if(isset($_POST["search_button"]))
        {        
            //create the sql statement 
            $sql_all= "SELECT * FROM testimonials";
            $result = mysqli_query($con,$sql_all);
             while ($row = mysqli_fetch_assoc($result)) 
                {
                echo   
                "<h4>Testimonial ID:</h4>". $row["testimonial_ID"].
                "<h4>Title:</h4> ".         $row["testimonial_title"].
                "<h4>Text:</h4> ".          $row["testimonial_text"].
                "<h4>Student ID:</h4>".     $row["student_ID"].
                "</br>";
                }

            mysqli_close($con);
        }

  include("footer.php");    
  ?>   
 </body>  
</html>

从数据库打印表格详细信息

0 个答案: