不显示 PHP 上的数据，但在工作台上 SQL 是正确的

Question

If I run the sql query on workbench, I can see the data. But on PHP there is nothing. I found this part doesn't work.

echo '<ul><li>'.$title1.'</li>';
echo '<li>'.$content1.'</li>';   
echo '<li>'.$answer1.'</li></ul>'; 

Does anyone know how to solve it?

                  $qtitle =$_GET['id'];

                  $sql ="select Q1.title, Q1.content, Q1.answer from questionset_question as Q2";
                  $sql .=" inner join question as Q1 on Q1.question_id=Q2.question_id inner join questionset as Q3";
                  $sql .=" on Q3.questionset_id=Q2.questionset_id where Q3.title='$qtitle'";

                  if ($result = $db->query($sql)) { 
                      while ($row = $result->fetch_assoc()) {   
                        $title1=$row['Q1.title'];
                        $content1=$row['Q1.content'];
                        $answer1=$row['Q1.answer'];

                        echo '<ul><li>'.$title1.'</li>';
                        echo '<li>'.$content1.'</li>';   
                        echo '<li>'.$answer1.'</li></ul>'; 
                      }

                      $result->free();
                    }
                    else {
                      echo "no result";
                    }

Answer 1

can you Try this way?

while ($row = $result->fetch_assoc()) {   
 $title1=$row['title'];
 $content1=$row['content'];
 $answer1=$row['answer']

Delete Q1 table name. Because the result in pdo returns without table name. MySQL workbench can list column name with table name.

Answer 2

In PHP or other web applications it is always good idea to see the contents of the variable before proceeding ahead and displaying the data

Here I would use print_r() function to see what $result contains inside.

if ($result = $db->query($sql)) { 

    //check the value in $result.. used only for debugging
    print_r($result);
    exit('debugging');

     while ($row = $result->fetch_assoc()) {   
         $title1 = $row['Q1.title'];
         $content1 = $row['Q1.content'];
         $answer1 = $row['Q1.answer'];
         ...
     }