沒有來自查詢結果的數據被回顯 - mysqli/php

Question

我想從查詢結果輸出數據到文本輸入和正常回聲，但它不輸出任何東西。我在PHP中沒有錯誤，我在mysqli代碼中做了什麼錯誤，它沒有迴應任何東西？

function ShowAssessment() 
    { 
     $sessiondetailsquery = " 
      SELECT SessionId, SessionName   
      FROM Session 
      WHERE 
      (SessionId = ?)"; 

     global $mysqli; 

     $sqlstmt=$mysqli->prepare($sessiondetailsquery); 


     $sqlstmt->bind_param("i",$_POST["session"]); 


     $sessiondetailsqrystmt=$mysqli->prepare($sessiondetailsquery); 
     // You only need to call bind_param once 
     $sessiondetailsqrystmt->bind_param("i",$_POST["session"]); 
     // get result and assign variables (prefix with db) 
     $sessiondetailsqrystmt->execute(); 
     $sessiondetailsqrystmt->bind_result($detailsSessionId,$detailsSessionName);?> 

     $sqlstmt->fetch(); 

     $sqlstmt->close(); 

     <h3>CHOSEN ASSESSMENT</h3> 
     <input type='text' id='currentId' name='Idcurrent' readonly='readonly' value='<?php $detailsSessionId; ?>' /></td> 
     <br> 
     <strong>Assessment:</strong> <?php echo $detailsSessionName; ?> 
     <?php 
    } 
    ?> 

Answer 1

你有2個bind_params在你的代碼中做同樣的事情。