Question

正如标题所说，当我用我的第二种形式（attendanceOut.php）创建一个带有mysqli查询的下拉列表时，我没有得到任何项目。我已经从我的attendance.php中的前一个下拉框中提取了一些信息，据我所知可以正常工作。我确定这与从INNER JOIN中提取的这个事实有关，但我无法找到我在语法上搞砸的地方。任何帮助将不胜感激。所以没有进一步的代码。

 <?php

    $course = $_GET["CourseName"];
    $startTime = $_GET["BeginTime"];
    $course = trim($course);
    $course = strip_tags($course);
    $startTime = trim($startTime);
    $startTime = strip_tags($startTime);
    ?>

    <html>
    <meta http-equiv="Content-Type" content="text/html"; charset=utf-8">
    <body>

    <?php

    $mydbconn = new mysqli("www.evilscriptmonkeys.com", "capstone", "capstone","attendance");

    if (mysqli_connect_errno())
    {
    printf("connection failed: %s\n", mysqli_connect_error());
    }

    if ($_GET['action'] == 'Create Form')
    {   
    //$result = mysqli_query($mydbconn, "SELECT * FROM student WHERE CourseName = ' " .$course."'");
    $c =$course;
    $s = $startTime;
    $sql = ("SELECT * FROM student WHERE CourseName = '$c' AND Time > '$s'");
    $result = $mydbconn->query($sql);
    printf ($course);
    printf ($startTime);
    echo '<table border="1">';

    $classFields = $result->fetch_fields();

    foreach ($classFields as $classData)
    {
        echo "<th> $classData->name </th>";
    }

    while ($row = $result->fetch_assoc())
    {
        echo "<tr>";
        foreach($row as $col=>$val)
        {
            echo "<td> $val </td>";
        }
        echo "</tr>";
    }

}
?>
<form method = "get" action="changeID.php">
Student Name: <select Name='UserName'>
<option value = "">---Select---</option>
<?php
if ($_GET['action'] == 'Edit Student ID')
{
    $c =$course;
    $result = mysqli_query($mydbconn, "SELECT s2.UserName 
                                       FROM student s1 
                                       INNER JOIN students s2 
                                       ON s1.UserName = s2.UserName 
                                       WHERE s1.CourseName = '$c'");
    while(($row = mysqli_fetch_assoc($result)))
    {
        echo "<option value=\" " . $row['s2.UserName'] . "\">". $row['s2.UserName'] . "</option>";
    }

}
?>
</select>
<input type="submit" name="action" value="Change" />
</body>
</html>

Answer 1

您可以使用别名，然后在数组中使用别名进行访问：

<?php
if ($_GET['action'] == 'Edit Student ID')
{
    $c =$course;
    $result = mysqli_query($mydbconn, "SELECT s2.UserName AS UserName
                                       FROM student s1 
                                       INNER JOIN students s2 
                                       ON s1.UserName = s2.UserName 
                                       WHERE s1.CourseName = '$c'");
    while(($row = mysqli_fetch_assoc($result)))
    {
        echo "<option value=\" " . $row['UserName'] . "\">". $row['UserName'] . "</option>";
    }

}
?>

希望得到这个帮助。

下拉列表没有显示mysqli查询中的任何项目

1 个答案: