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[英]How do I use php to upload and rename multiple files using a dropdown menu for the file names?
[英]How do I put names from table into dropdown menu?
我试图弄清楚为什么下拉列表中没有显示老师列表
include"teacher.php"
<body onload="displayDate()">
<img src="Raw Pictures/Header.jpg" style="width:100% ; height:15%">
<img src="Raw Pictures/green.png" style="width:8%; height:11%; position:absolute; top:4% ;left: 3%">
<br><br><br><div>
Teacher:
<select>
<option>echo $row</option>
</select>
</body>
</html>
然后是Teacher.php
<?php
mysql_connect('localhost', 'root', 'password');
mysql_select_db('teacher_account');
$sql = "SELECT facultyname FROM subj_eva";
$result = mysql_query($sql);
while ($row = mysql_fetch_array($result)) {}
?>
我不知道如何显示教师从数据库到下拉框中? 我如何解决它?
您可能正在寻找类似的东西:
<?php
mysql_connect('localhost', 'root', 'password');
mysql_select_db('teacher_account');
$sql = "SELECT teachername,facultyname FROM subj_eva";
$result = mysql_query($sql);
?>
<html>
<body onload="displayDate()">
<img src="Raw Pictures/Header.jpg"
style="width:100% ; height:15%">
<img src="Raw Pictures/green.png"
style="width:8%; height:11%; position:absolute; top:4% ;left: 3%">
<div>
Teacher:
<select>
<?php while ($row = mysql_fetch_array($result)) { ?>
<option value="<?= $row['teachername'] ?>">
<?= $row['teachername'] ?> (<?= $row['facultyname'] ?>)
</option>
<?php } ?>
</select>
</body>
</html>
当然,代码可以分成文件,为便于阅读,我将其捆绑在一起。 将这些样式规则移动到单独的css
文件中,而不是使用那些难看的内联样式可能是有意义的。
附带说明:您正在使用过时和不建议使用的mysql
扩展。 您应该切换到mysqli
或PDO
并了解“预备语句”的安全性优势。 另外,您不应将root
帐户用于正常的数据库使用。 创建一个特权较低的帐户,仅将根帐户用于管理任务。
<?php
mysql_connect('localhost', 'root', 'password');
mysql_select_db('teacher_account');
$sql = "SELECT facultyname FROM subj_eva";
$result = mysql_query($sql);
while ($result_array = mysql_fetch_array($result, MYSQL_ASSOC) ) {
$row = $row .'<option>'. $result_array["facultyname"] .'</option>';
}
?>
和PHP生成HTML文件
<?php
include"teacher.php";
?>
<body onload="displayDate()">
<img src="Raw Pictures/Header.jpg" style="width:100% ; height:15%">
<img src="Raw Pictures/green.png" style="width:8%; height:11%; position:absolute; top:4% ;left: 3%">
<br><br><br><div>
Teacher:
<select>
<?php
echo $row;
?>
</select>
</body>
</html>
尝试这个
$sql = "SELECT facultyname FROM subj_eva";
$result = mysql_query($sql);
echo "<select>";
while ($row = mysql_fetch_array($result))
{
echo "<option>$row['column']</option>";}
echo "</select>";
?>
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