Php 下拉列表不显示mysqli查询中的任何项目
正如标题所说，当我用第二种形式（attendanceOut.php）的mysqli查询创建下拉列表时，我没有得到任何项目。我已经从my Attention.php的上一个下拉框中提取了一些信息，据我所知，该下拉框工作正常。我肯定这是因为这是从一个内部连接中提取出来的，但是我找不到我把语法弄乱的地方。任何帮助都将不胜感激。因此，无需进一步修改一些代码Php 下拉列表不显示mysqli查询中的任何项目,php,mysql,drop-down-menu,inner-join,Php,Mysql,Drop Down Menu,Inner Join,正如标题所说，当我用第二种形式（attendanceOut.php）的mysqli查询创建下拉列表时，我没有得到任何项目。我已经从my Attention.php的上一个下拉框中提取了一些信息，据我所知，该下拉框工作正常。我肯定这是因为这是从一个内部连接中提取出来的，但是我找不到我把语法弄乱的地方。任何帮助都将不胜感激。因此，无需进一步修改一些代码 <?php $course = $_GET["CourseName"]; $startTime = $_GET["Begi
<?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'] . "\">". $row['s2.UserName'] . "</option>";
}
}
?>
</select>
<input type="submit" name="action" value="Change" />
</body>
</html>
您可以使用别名，然后在阵列中使用它进行访问：
<?php
if ($_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>";
}
}
?>
$row['s2.UserName']
到$row['UserName']
谢谢这就是问题所在。我可能会建议编辑您的数据库连接详细信息（如果这些是合法的）