【发布时间】:2013-01-01 04:48:24
【问题描述】:
当用户从问题下拉菜单中选择All 并在下方输出时,我想显示下拉菜单中的所有问题。问题是它没有这样做,更糟的是,它给了我未定义的偏移错误,说明:
Notice: Undefined offset: ... in .... on line 605
第 605 行是:
echo '<p><strong>Question:</strong> ' .htmlspecialchars($arrQuestionNo[$key]). ': ' .htmlspecialchars($arrQuestionContent[$key]). '</p>' . PHP_EOL;
如果用户选择All 选项,我的问题是如何修复错误并显示所有问题?
我有一个演示,你可以通过:DEMO
按照以下步骤操作:
- 在模块下拉菜单中,选择系统策略并提交
- 当评估下拉菜单出现时,选择 POKUB1 并提交
- 您将看到学生和问题下拉菜单。如果打开下拉菜单,您会看到有 3 个学生和 2 个问题。请选择一个学生和
All问题并提交。当我真的想在此处显示所有问题的详细信息时,您会在这里看到错误。
代码:
问题下拉菜单:
<select name="question" id="questionsDrop">
<option value="0">All</option>
<option value=23">1</option>
<option value=32">1</option>
</select>
以下是根据从问题下拉菜单中选择的选项确定显示的代码。
function StudentAnswers()
{
$selectedstudentanswerqry = "
SELECT
sa.StudentId, StudentAlias, StudentForename, StudentSurname, q.SessionId,
QuestionNo, QuestionContent, o.OptionType, q.NoofAnswers,
GROUP_CONCAT( DISTINCT Answer ORDER BY Answer SEPARATOR ',' ) AS Answer, r.ReplyType, QuestionMarks,
GROUP_CONCAT(DISTINCT StudentAnswer ORDER BY StudentAnswer SEPARATOR ',') AS StudentAnswer, ResponseTime, MouseClick, StudentMark
FROM Student st
INNER JOIN Student_Answer sa ON (st.StudentId = sa.StudentId)
INNER JOIN Student_Response sr ON (sa.StudentId = sr.StudentId)
INNER JOIN Question q ON (sa.QuestionId = q.QuestionId)
INNER JOIN Answer an ON q.QuestionId = an.QuestionId
LEFT JOIN Reply r ON q.ReplyId = r.ReplyId
LEFT JOIN Option_Table o ON q.OptionId = o.OptionId
";
// Initially empty
$where = array('q.SessionId = ?');
$parameters = array($_POST["session"]);
$parameterTypes = 'i';
// Check whether a specific question was selected
$p_question = empty($_POST["question"])?'':$_POST["question"];
switch($p_question){
case 0:
//dont' add where filters
break;
default:
$where[] = 'q.QuestionId = ?';
$parameters[] .= $_POST["question"];
$parameterTypes .= 'i';
}
// If we added to $where in any of the conditionals, we need a WHERE clause in
// our query
if(!empty($where)) {
$selectedstudentanswerqry .= ' WHERE ' . implode(' AND ', $where);
global $mysqli;
$selectedstudentanswerstmt=$mysqli->prepare($selectedstudentanswerqry);
// You only need to call bind_param once
if (count($where) == 1) {
$selectedstudentanswerstmt->bind_param($parameterTypes, $parameters[0]);
}
else if (count($where) == 2) {
$selectedstudentanswerstmt->bind_param($parameterTypes, $parameters[0], $parameters[1]);
}
}
$selectedstudentanswerqry .= "
GROUP BY sa.StudentId, q.QuestionId
ORDER BY StudentAlias, q.SessionId, QuestionNo
";
// get result and assign variables (prefix with db)
$selectedstudentanswerstmt->execute();
$selectedstudentanswerstmt->bind_result($detailsStudentId,$detailsStudentAlias,$detailsStudentForename,$detailsStudentSurname,$detailsSessionId,$detailsQuestionNo,
$detailsQuestionContent,$detailsOptionType,$detailsNoofAnswers,$detailsAnswer,$detailsReplyType,$detailsQuestionMarks,$detailsStudentAnswer,$detailsResponseTime,
$detailsMouseClick,$detailsStudentMark);
$selectedstudentanswerstmt->store_result();
$selectedstudentanswernum = $selectedstudentanswerstmt->num_rows();
$question = array();
while ($selectedstudentanswerstmt->fetch()) {
$arrQuestionNo = array();
$arrQuestionContent = array();
$arrQuestionNo[ $detailsStudentId ] = $detailsQuestionNo;
$arrQuestionContent[ $detailsStudentId ] = $detailsQuestionContent;
$questions[] = $arrQuestionNo;
$questions[] = $arrQuestionContent;
}
$selectedstudentanswerstmt->close();
?>
...........................................................................................
<h2>STUDENT'S ANSWERS</h2>
<?php
foreach ($questions as $key=>$question) {
echo '<p><strong>Question:</strong> ' .htmlspecialchars($arrQuestionNo[$key]). ': ' .htmlspecialchars($arrQuestionContent[$key]). '</p>' . PHP_EOL;
}
}
?>
更新:
学生表结构:
CREATE TABLE `Student` (
`StudentId` int(10) NOT NULL AUTO_INCREMENT,
`StudentForename` varchar(25) NOT NULL,
`StudentSurname` varchar(25) NOT NULL,
`StudentAlias` varchar(15) NOT NULL,
`StudentEmail` varchar(50) NOT NULL,
`StudentUsername` varchar(20) NOT NULL,
`StudentPassword` varchar(50) NOT NULL,
`StudentDOB` date NOT NULL,
`Year` int(2) NOT NULL,
`CourseId` int(6) NOT NULL,
`Active` tinyint(1) NOT NULL DEFAULT '1',
PRIMARY KEY (`StudentId`),
KEY `FK_Course` (`CourseId`)
) ENGINE=InnoDB AUTO_INCREMENT=41 DEFAULT CHARSET=utf8
问题表结构:
CREATE TABLE `Question` (
`QuestionId` int(10) NOT NULL AUTO_INCREMENT,
`SessionId` int(10) NOT NULL,
`QuestionNo` int(3) NOT NULL,
`QuestionContent` varchar(5000) NOT NULL,
`NoofAnswers` int(2) NOT NULL,
`ReplyId` int(1) NOT NULL,
`QuestionMarks` int(4) NOT NULL,
`OptionId` int(2) NOT NULL,
PRIMARY KEY (`QuestionId`)
) ENGINE=InnoDB AUTO_INCREMENT=357 DEFAULT CHARSET=utf8
【问题讨论】:
-
你不是在一小时前就问过这个问题了吗?并且...尽量复制更少的代码,更重要的是sn-ps,没有人愿意阅读几页代码来查找(例如)错字。
-
@VladPreda 这是相似的代码但不同的问题。我被告知要分开我的问题,所以我在一个问题中问一个问题,在这个问题中问这个问题。我会减少代码,我只是想让大家知道我的代码结构
-
在 Helios 计算与工程学院,他们需要教您如何调试和理解您编写的代码。您正在解释问题,就好像您是网站用户(“我选择”,“它显示”),而不是编写代码的程序员。程序员必须能够遵循程序逻辑并将实际结果与预期结果进行比较。在变量值和控制流方面,而不是像“它没有出现”这样的文学术语。
-
@user1914374 您需要重构代码以使其更易于调试(将 DAL、逻辑和视图相互分离,忘记过程脚本并使用 OO 结构(尤其是 single responsibility principle,使用已经存在的设计模式,如MVC)。顺便说一句,如果你只想隐藏通知,你可以简单地检查是否定义了索引:
...htmlspecialchars((isset($arrQuestionNo[$key])?$arrQuestionNo[$key]:''))...