Question

我创建了一个网站，用于帮助教授ECDL。在一个页面上，我有一个教师概述，它使用php和SQLi来取得学生的进度，并根据主题将其放入表中。Overview Page

我有3个主题，我使用标签进行导航，为了避免冗余，我已经把代码填充到一个php文件中，并在html中使用包含vaiables。第一个主题工作正常，但第二个和第三个主题是课程文本和学生名称被填充，但进度返回false，警告：mysqli_fetch_row（）期望参数1为mysqli_result，给定布尔值。

我试图取消设定变量，免费结果等无济于事。用户的第二个主题数据库表当前是空的，但第三个不是。 pres DB table如果我不使用inlcude文件，它们将有效，但我希望将来保持整洁/易于编辑。

Overview.php

    <ul class="tab">
     <li><a href="#" class="tablinks" onclick="openLesson(event, '1')">Word Processing</a></li>
     <li><a href="#" class="tablinks" onclick="openLesson(event, '2')">Speedsheets</a></li>
     <li><a href="#" class="tablinks" onclick="openLesson(event, '3')">Presentation</a></li>
</ul>
<div id="1" class="tabcontent" style="padding-top:0; max-height:2000px">
    <div>
        <?php 
            $mod = "word";
            $sec = "wp";

            include("../../SCRIPT/PHP/progOver.php");

        ?>
    </div>
</div>
<div id="2" class="tabcontent" style="padding-top:0; max-height:2000px">
    <div>
        <?php 
            $mod = "excel";
            $sec = "ss";

            include("../../SCRIPT/PHP/progOver.php");   

        ?>
    </div>
</div>
<div id="3" class="tabcontent" style="padding-top:0; max-height:2000px">
<div>
        <?php 
            $mod = "pres";
            $sec = "pp";

            include("../../SCRIPT/PHP/progOver.php");

        ?>
    </div>
</div>

progOver.php

$sql_text = mysqli_query($con,"SELECT * FROM ".$sec."_text ORDER BY id ASC");
$sql_users ="SELECT username, f_name, s_name FROM user WHERE class ='".$class."' ORDER BY s_name ASC";

$x = 1;
$time = "";

if($result = mysqli_query($con, $sql_users)){
    echo"<div class ='xyz'><table class='progress'>";
    echo"<thead >\n<tr>\n";
    echo"<th class='headcol' style='width:100px;height:250px;'></th>\n";
        while($row = mysqli_fetch_array($sql_text)){
            echo"<th class ='vert' ><div class='vert'>".$row["name"]."</div></th>\n";
        }
        //print_r($result);
        echo "</tr>\n</thead>\n<tbody>\n";
        while($row_user = mysqli_fetch_row($result)){
            echo "<tr>\n";
            $sql_rag = "SELECT ".$mod.".".$sec."_lesson, ".$mod.".rag, ".$sec."_text.id, ".$mod.".comp_on FROM ".$mod." INNER JOIN ".$sec."_text ON ".$mod.".".$sec."_lesson = ".$sec."_text.".$sec."_lesson WHERE username = '".$row_user[0]."' ORDER BY ".$mod.".username, ".$sec."_text.id ASC";
            $result_rag = mysqli_query($con, $sql_rag);
            echo '<th class="headcol">'.$row_user[1].' '.$row_user[2].'</th>';
            //print_r($result_rag);
            if($result_rag){
                while($row_rag = mysqli_fetch_row($result_rag)){ //Error here
                    $time = $row_rag[3];
                    if($x == $row_rag[2]){
                        if($row_rag[1] == "Green"){
                            echo "<td class='long'style='background-color:green;' alt=".$time." title=".$time."></td>\n";
                            $x++;
                        }else if($row_rag[1] == "Red"){
                            echo "<td class='long' style='background-color:red;' alt=".$time." title=".$time."></td>\n";
                            $x++;
                        }if($row_rag[1] == "Amber"){
                            echo "<td class='long' style='background-color:orange;' alt=".$time." title=".$time."></td>\n";
                            $x++;
                        }
                    }else{
                        echo"<td class='long'>X</td>\n";
                        $x = $x+2;
                    }
                }
            }
        echo "</tr>\n";
        $x=1;
        }
    echo"</table>\n</div>\n";
    mysqli_free_result($result);

}

SQLi没有结果，代码冗余

0 个答案: