表单中包含MySQL的HTML下拉框
我有PHP代码可以工作,但我没有运气将其转换为HTML表单。有什么建议吗表单中包含MySQL的HTML下拉框,mysql,Mysql,我有PHP代码可以工作,但我没有运气将其转换为HTML表单。有什么建议吗 <?php $db_host = "localhost"; $db_username = "combsb_combsb"; $db_pass = "pat60086"; $db_name = "combsb_sample"; @mysql_connect ("$db_host", "$db_username", "$db_pass") or die ("Could not connect to MySQL
<?php
$db_host = "localhost";
$db_username = "combsb_combsb";
$db_pass = "pat60086";
$db_name = "combsb_sample";
@mysql_connect ("$db_host", "$db_username", "$db_pass") or die ("Could not connect to MySQL");
@mysql_select_db("$db_name") or die ("No Database");
Echo"Successful Connection";
$sql = "SELECT compname FROM Crew";
$result = mysql_query($sql);
echo "<select name='compname'>";
while ($row = mysql_fetch_array($result)) {
echo "<option value='" . $row['compname'] . "'>" . $row['compname'] . "</option>";
}
echo "</select>";
?>
首先,我要确保在PHP中添加HTML样板文件。那么我想你指的是mysql\u fetch\u行而不是mysql\u fetch\u数组。您的最终文件应类似于:
<!DOCTYPE html>
<html>
<head></head>
<body>
<?php
// *Cut out for brevity, remember to paste back in*
$result = mysql_query($sql);
?>
<form action="" method="POST">
<?php
echo "<select name='compname'>";
while ($row = mysql_fetch_row($result)) {
echo "<option value='" . $row['compname'] . "'>" . $row['compname'] . "</option>";
}
echo "</select>";
?>
</form>
</body>
</html>
下次发布您的输出/错误,或者如果这不起作用您希望表中的db项填充选择框?是的,建议不要使用MySQL库访问数据:)显示您的错误。告诉我们您实际遇到了什么问题。因为你“运气不好”而寻求一般性的建议不会给任何人带来太多的进展。如果你用具体问题发帖,你会得到更好的答案。