PHP代码未成功运行

Question

我的问题很简单。我在这里编写的代码绝对不会在网页上产生任何输出。我整天都在做这件事，我确信这是一件非常简单的事情，我错过了一个白痴。所以我很吸引你善良的新鲜眼睛！如果有人能找出这不起作用的原因，我将不胜感激。

前提：

这是一个决策树在线调查，具有以下条件：如果用户已经开始调查，它将在数据库中找到他们，找到他们最后回答的问题并显示下一个。但如果他们还没有开始，它会显示第一个问题。

所有调查问题以及决策树逻辑都保存在数据库中（例如，如果用户为问题 1 选择选项 2，他们将被引导至问题 3，而不是问题 2）。

请假设目前，我是直接从数据库中更新相关信息，而不是在网站上自动更新。

谢谢:)

PHP：

    <?php
//Find the latest question reached by the user for display on the page
$sql = mysql_query("SELECT QuestionNumberReached FROM User WHERE EmailAddress = '***'");
$sqlCount = mysql_num_rows($sql);
if ($sqlCount > 0) {
    while ($row = mysql_fetch_array($sql)) {
        $QuestionNumberReached = $row["QuestionNumberReached"];
    }
}
?>

<?php
//Find the last question answered by the user from the database
$StartedQuery = mysql_query("SELECT LastQuestionAnswered FROM User WHERE EmailAddress = '***'");
//Count the number of rows that the query produces
$StartedQueryCount = mysql_num_rows($StartedQuery);
//If data is found, whether it be a number or null, define the value
if ($StartedQueryCount > 0) {
    while ($row = mysql_fetch_array($sql)) {
        $LastQuestionAnswered = $row["LastQuestionAnswered"];
        //If the field has a value and is not null, find the next question from the database
        if (!empty($LastQuestionAnswered)) {
            //Find the User's ID and the ID of the last question answered
            $sqlA = mysql_query("SELECT PKID, LastQuestionAnswered FROM User WHERE EmailAddress = '***'");
            //If the operation produces an error, output an error message
            if (!$sqlA) {
                die('Invalid query for SQLA: ' . mysql_error());
            }
            //Count the number of rows output
            $sqlACount = mysql_num_rows($sqlA);
            //If rows exist, define the values
            if ($sqlACount > 0) {
                while ($row = mysql_fetch_array($sqlA)) {
                    $sqlAPKID = $row["PKID"];
                    $sqlALastQuestionAnswered = $row["LastQuestionAnswered"];
                }
            }
            //Find the answer given by the user to the last answered question
            $sqlB = mysql_query("SELECT Answer FROM Responses WHERE User = $sqlAPKID");
            //If the operation produces an error, output an error message
            if (!$sqlB) {
                die('Invalid query for SQLB: ' . mysql_error());
            }
            //Count the number of rows output
            $sqlBCount = mysql_num_rows($sqlB);
            //If rows exist, define the values
            if ($sqlBCount > 0) {
                while ($row = mysql_fetch_array($sqlB)) {
                    $sqlBAnswer = $row["Answer"];
                }
            }
            //Find the number of the next question to be answered based on the user's previous answer and the question they answered
            $sqlC = mysql_query("SELECT NextQuestion FROM Answers WHERE QuestionNumber = $sqlALastQuestionAnswered AND PKID = $sqlBAnswer");
            //If the operation produces an error, output an error message
            if (!$sqlC) {
                die('Invalid query for SQLC: ' . mysql_error());
            }
            //Count the number of rows output
            $sqlCCount = mysql_num_rows($sqlC);
            //If rows exist, define the values
            if ($sqlCCount > 0) {
                while ($row = mysql_fetch_array($sqlC)) {
                    $sqlCNextQuestion = $row["NextQuestion"];
                }
            }
            //Find the question text pertaining to the ID of the next question that needs to be answered
            $sqlD = mysql_query("SELECT QuestionText FROM Questions WHERE PKID = $sqlCNextQuestion");
            //If the operation produces an error, output an error message
            if (!$sqlD) {
                die('Invalid query for SQLD: ' . mysql_error());
            }
            //Count the number of rows output
            $sqlDCount = mysql_num_rows($sqlD);
            //If rows exist, define the values
            if ($sqlDCount > 0) {
                while ($row = mysql_fetch_array($sqlD)) {
                    $SurveyStartedQuestionText = $row["QuestionText"];
                }
            }
            //Set a string of information that will show the question number and question text as appropriate
            $ToDisplay = '' . $QuestionNumberReached . ': ' . $SurveyStartedQuestionText . '<br /><br />Answer Text Here';
        //If the value for QuestionNumberReached is null, the user has not started the survey
        } else if (empty($LastQuestionAnswered)) {
            //Find the question text of the first question in the survey
            $sql3 = mysql_query("SELECT QuestionText FROM Questions WHERE PKID IN (SELECT FirstQuestion FROM Batch WHERE BatchNumber IN (SELECT BatchNumber FROM User WHERE EmailAddress = '***'))");
            //Count the number of rows output
            $sql3Count = mysql_num_rows($sql3);
            //If rows exist, define the values
            if ($sql3Count > 0) {
                while ($row = mysql_fetch_array($sql3)) {
                    $SurveyNotStartedQuestionText = $row["QuestionText"];
                }
            }
            //Set a string of information that will show the question number and question text as appropriate
            $ToDisplay = '' . $QuestionNumberReached . ': ' . $SurveyNotStartedQuestionText . '<br /><br />Answer Text Here';
        }
    }
}
?>

HTML：

<body>

<?php
// Display the concatenated information that has been previously defined
echo $ToDisplay;
?>

</body>

Answer 1

这一点：

if ($StartedQueryCount > 0) {

可能评估为 false，并且没有匹配的 else 标记添加内容。 尝试改变：

    }
?>

与：

    }
    else {
        $ToDisplay = 'Error: no rows found to display!';
    }
?>

编辑：

还有，这一点：

} else if (empty($LastQuestionAnswered)) {

可以换成更易读的：

} else {

因为它做的事情完全一样。 在你的 while 循环中，你不断地重新定义 $ToDisplay，我认为这是想要的行为？否则初始化顶部的变量（在 while() 循环之前），如下所示：

$ToDisplay = '';

并将循环内的赋值更改为串联，如下所示：

$ToDisplay = 'text assignment';

收件人：

$ToDisplay .= 'text concat'; // look at the dot before =

Answer 2

感谢您的所有帮助！非常感谢大家抽出宝贵的时间。

我终于意识到出了什么问题……

在我的 PHP 代码的第 18 行，我有以下内容：

while ($row = mysql_fetch_array($sql)) {

当然应该是这样的：

while ($row = mysql_fetch_array($StartedQuery)) {

基本上我是从错误的查询中调用行。我因此而感到凝块！

再次感谢大家:)