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我无法在各自的文本输入中显示 SessionName、SessionDate 和 SessionTime。应该发生的是用户应该从下拉菜单中选择一个会话（评估）。现在，当他们提交下拉菜单时，应在其文本输入中显示 Session 的详细信息，即 SessionName、SessionDate 和 SessionTime。但相反，我收到未定义的变量错误，这些错误如下：

注意：未定义的变量：dbSessionName in ...第 243 行

注意：未定义的变量：dbSessionDate in ... 第 244 行

注意：未定义的变量：dbSessionTime in ... 第 245 行

如何让 SessionName、SessionTime 和 SessionDate 显示在它们各自的文本输入中？

下面是代码：

$sessionquery = "
SELECT SessionId, SessionName, SessionDate, SessionTime, ModuleId
FROM Session
WHERE
(ModuleId = ?)
ORDER BY SessionDate, SessionTime 
";

$sessionqrystmt=$mysqli->prepare($sessionquery);
// You only need to call bind_param once
$sessionqrystmt->bind_param("s",$_POST['modules']);
// get result and assign variables (prefix with db)

$sessionqrystmt->execute(); 

$sessionqrystmt->bind_result($dbSessionId,$dbSessionName,$dbSessionDate,$dbSessionTime, $dbModuleId);

$sessionqrystmt->store_result();

$sessionnum = $sessionqrystmt->num_rows();   

if($sessionnum ==0){
echo "<p>Sorry, You have No Assessments under this Module</p>";
} else { 

$sessionHTML = '<select name="session" id="sessionsDrop">'.PHP_EOL;
$sessionHTML .= '<option value="">Please Select</option>'.PHP_EOL;           

while ( $sessionqrystmt->fetch() ) {
    $sessionHTML .= sprintf("<option value='%s'>%s - %s - %s</option>", $dbSessionId, $dbSessionName, $dbSessionDate, $dbSessionTime) . PHP_EOL;  
}

$sessionHTML .= '</select>';

$assessmentform = "<form action='".htmlentities($_SERVER['PHP_SELF'])."' method='post' onsubmit='return sessionvalidation();'>
<p>Assessments: {$sessionHTML} </p>
<p><input id='sessionSubmit' type='submit' value='Submit Assessment' name='sessionSubmit' /></p>  
<div id='sessionAlert'></div>    
</form>";

echo $assessmentform;

}

}

if (isset($_POST['sessionSubmit'])) {   

$currentsession = "form action='".htmlentities($_SERVER['PHP_SELF'])."' method='post'>
<p>Current Assessment's Date/Start Time:</p>
<p>Assessment: <input type='text' id='currentAssessment' name='Assessmentcurrent' readonly='readonly' value='{$dbSessionName}'/> </p> //Line 243
<p>Date: <input type='text' id='currentDate' name='Datecurrent' readonly='readonly' value='{$dbSessionDate}'/> </p> //Line 244
<p>Start Time: <input type='text' id='currentTime' name='Timecurrent' readonly='readonly' value='{$dbSessionTime}'/> </p> //Line 245
</form>
";  

echo $currentsession;

    }

更新：

下面的代码可以做到吗：

if (isset($_POST['sessionSubmit'])) {

    $sessionquery = "
    SELECT SessionId, SessionName, SessionDate, SessionTime, ModuleId
    FROM Session
    WHERE
    (ModuleId = ?)
    ORDER BY SessionDate, SessionTime 
    ";

    $sessionqrystmt=$mysqli->prepare($sessionquery);
    // You only need to call bind_param once
    $sessionqrystmt->bind_param("s",$_POST['modules']);
    // get result and assign variables (prefix with db)

    $sessionqrystmt->execute(); 

    $sessionqrystmt->bind_result($dbSessionId,$dbSessionName,$dbSessionDate,$dbSessionTime, $dbModuleId);

    $sessionqrystmt->store_result();

$currentsession = "<form action='".htmlentities($_SERVER['PHP_SELF'])."' method='post'>
<p>Current Assessment's Date/Start Time:</p>
<p>Assessment: <input type='text' id='currentAssessment' name='Assessmentcurrent' readonly='readonly' value='{$dbSessionName}'/> </p>
<p>Date: <input type='text' id='currentDate' name='Datecurrent' readonly='readonly' value='{$dbSessionDate}'/> </p>
<p>Start Time: <input type='text' id='currentTime' name='Timecurrent' readonly='readonly' value='{$dbSessionTime}'/> </p>
<input type='hidden' id='hiddenId' name='hiddenAssessment' value='{$dbSessionId}'/>
</form>
";  

echo $currentsession;

    }

我需要 store_result(); 在这个情况下？

4

1 回答 1

0

您实际上并没有使用发布的值session从数据库中检索选定的会话。

您显示的 sql 用于生成select框并session在发布的表单中选择，您需要在

if (isset($_POST['sessionSubmit'])) {
  ...
}

部分使用 获取先前选择的会话$_POST['session']，可能类似于：

$sessionquery = "
SELECT SessionId, SessionName, SessionDate, SessionTime, ModuleId
FROM Session
WHERE
(SessionId = ?)
";
....
$sessionqrystmt->bind_param("i",$_POST['session']);    // assuming the session ID is an integer
于 2012-10-31T22:46:49.893 回答