php不从数据库中获取数据

时间:2013-07-18 20:29:32

标签: php jquery mysql

session_start();
require_once("scripts/connect_db.php");
   $arrCount = "";
         if(isset($_GET['question'])){
              $question = preg_replace('/[^0-9]/', "", $_GET['question']);
           $output = "";
          $answers1 = "";
          $answers2 = "";
          $answers3 = "";
          $answers4 = "";
            $q = "";
        $sql = mysql_query("SELECT id FROM questions");
        $numQuestions = mysql_num_rows($sql);
       if(!isset($_SESSION['answer_array']) || $_SESSION['answer_array'] < 1){
    $currQuestion = "1";
}else{
    $arrCount = count($_SESSION['answer_array']);
}
if($arrCount > $numQuestions){
    unset($_SESSION['answer_array']);
    header("location: index.php");
    exit();
}
if($arrCount >= $numQuestions){
    echo 'finished|<p>There are no more questions. Please enter your first and last name and click next</p>
            <form action="userAnswers.php" method="post">
            <input type="hidden" name="complete" value="true">
            <input type="text" name="username">
            <input type="submit" value="Finish">
            </form>';
    exit();
}
$singleSQL = "SELECT * FROM questions WHERE id='$question' LIMIT 1 ";
$result=mysql_query($singleSQL);
if($singleSQL){
    die( mysql_error());
}
    while($row = mysql_fetch_array($result)){
        $id = $row['id'];
        $thisQuestion = $row['question'];
        $type = $row['type'];
        $question_id = $row['question_id'];
        $q = '<h2>'.$thisQuestion.'</h2>';
        $sql2 = mysql_query("SELECT * FROM answers WHERE question_id='$question' ORDER BY rand()");
        while($row2 = mysql_fetch_array($sql2)){
            $answer1 = $row2['answer1'];
            $answer2 = $row2['answer2'];
            $answer3 = $row2['answer3'];
            $answer4 = $row2['answer4'];
            $correct = $row2['correct'];
            $answers1 .= '<label style="cursor:pointer;"><input type="radio" name="rads" value="'.$correct.'">'.$answer1.'</label> 
            <input type="hidden" id="qid" value="'.$id.'" name="qid"><br /><br />
            ';
            $answers2 .= '<label style="cursor:pointer;"><input type="radio" name="rads" value="'.$correct.'">'.$answer2.'</label> 
            <input type="hidden" id="qid" value="'.$id.'" name="qid"><br /><br />
            ';
            $answers3 .= '<label style="cursor:pointer;"><input type="radio" name="rads" value="'.$correct.'">'.$answer3.'</label> 
            <input type="hidden" id="qid" value="'.$id.'" name="qid"><br /><br />
            ';
            $answers4 .= '<label style="cursor:pointer;"><input type="radio" name="rads" value="'.$correct.'">'.$answer4.'</label> 
            <input type="hidden" id="qid" value="'.$id.'" name="qid"><br /><br />
            ';
        }
        $output = ''.$q.','.$answers1.','.$answers2.','.$answers3.','.$answers4.',<span id="btnSpan"><button onclick="post_answer()">Submit</button></span>';
        echo $output;
       }
    }

它是一个关于在线考试系统的项目我能够插入数据但是在获取它无法显示它只是给出错误时undefined plz帮帮我 我应该显示问题和四个选项并检查答案并给我标记 我的HTML就像这样

<body>
            <p id="question"></p>
                      <table>
                                   <tr>
                           <td id="answer1"><td>
                                   </tr>
                                  <tr>
                          <td id="answer2"></td>
                               </tr>
                                  <tr>
                        <td id="answer3"></td>
                               </tr>
                                    <tr>
                          <td id="answer4"></td>
                                     <tr>
                                                   </table>

1 个答案:

答案 0 :(得分:2)

你的脚本总是被die()停止:

if($singleSQL){
    die( mysql_error());
}