php 如何使用数据库中的值填充 HTML 下拉列表
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How to populate HTML dropdown list with values from database
提问by Bernard
as part of a HTML form I am creating I would like to have a dropdown list which will list all the usernames in my database.
作为我正在创建的 HTML 表单的一部分，我想要一个下拉列表，该列表将列出我的数据库中的所有用户名。
I thought the following code would do the trick but the dropdown list is empty - could someone assist me in what i'm doing wrong? Thanks.
我认为以下代码可以解决问题，但下拉列表是空的 - 有人可以帮助我解决我做错了什么吗？谢谢。
<tr>
<td>Owner</td>
<td>
<select name="owner">
<?php
$sql = mysqli_query($connection, "SELECT username FROM users");
while ($row = $sql->fetch_assoc()){
?>
<option value="owner1"><?php echo $row['username']; ?></option>
<?php
// close while loop
}
?>
</td>
</tr>
回答by Christofer Eliasson
My guess is that you have a problem since you don't close your select-tag after the loop. Could that do the trick?
我的猜测是你有问题，因为你没有在循环后关闭你的选择标签。这能行吗？
<select name="owner">
<?php
$sql = mysqli_query($connection, "SELECT username FROM users");
while ($row = $sql->fetch_assoc()){
echo "<option value=\"owner1\">" . $row['username'] . "</option>";
}
?>
</select>
回答by Imon
Below code is nice.. It was given by somebody else named aaronbd in this forum
下面的代码很好..它是由本论坛中名为 aaronbd 的其他人提供的
<?php
$conn = new mysqli('localhost', 'username', 'password', 'database')
or die ('Cannot connect to db');
$result = $conn->query("select id, name from table");
echo "<html>";
echo "<body>";
echo "<select name='id'>";
while ($row = $result->fetch_assoc()) {
unset($id, $name);
$id = $row['id'];
$name = $row['name'];
echo '<option value="'.$id.'">'.$name.'</option>';
}
echo "</select>";
echo "</body>";
echo "</html>";
?>
回答by Kris C
I'd suggest following a few debugging steps.
我建议遵循一些调试步骤。
First run the query directly against the DB. Confirm it is bringing results back. Even with something as simple as this you can find you've made a mistake, or the table is empty, or somesuch oddity.
首先直接对数据库运行查询。确认它正在返回结果。即使是这样简单的事情，你也会发现你犯了一个错误，或者桌子是空的，或者一些奇怪的事情。
If the above is ok, then try looping and echoing out the contents of $row just directly into the HTML to see what you've getting back in the mysql_query - see if it matches what you got directly in the DB.
如果以上没有问题，那么尝试循环并将 $row 的内容直接回显到 HTML 中，以查看您在 mysql_query 中返回的内容 - 查看它是否与您直接在数据库中获得的内容匹配。
If your data is output onto the page, then look at what's going wrong in your HTML formatting.
如果您的数据输出到页面上，那么请查看您的 HTML 格式中出了什么问题。
However, if nothing is output from $row, then figure out why the mysql_query isn't working e.g. does the user have permission to query that DB, do you have an open DB connection, can the webserver connect to the DB etc [something on these lines can often be a gotcha]
但是，如果没有任何输出$row，那么找出 mysql_query 不工作的原因，例如用户是否有权查询该数据库，您是否有一个开放的数据库连接，网络服务器是否可以连接到数据库等 [这些行上的某些内容通常可能是一个陷阱]
Changing your query slightly to
将您的查询稍微更改为
$sql = mysql_query("SELECT username FROM users") or die(mysql_error());
may help to highlight any errors: php manual
可能有助于突出任何错误：php 手册
回答by user2806221
<select name="owner">
<?php
$sql = mysql_query("SELECT username FROM users");
while ($row = mysql_fetch_array($sql)){
echo "<option value=\"owner1\">" . $row['username'] . "</option>";
}
?>
</select>
回答by rashedcs
<?php
$query = "select username from users";
$res = mysqli_query($connection, $query);
?>
<form>
<select>
<?php
while ($row = $res->fetch_assoc())
{
echo '<option value=" '.$row['id'].' "> '.$row['name'].' </option>';
}
?>
</select>
</form>
回答by tim3in
Use OOP concept instead. Create a class with function
请改用 OOP 概念。用函数创建一个类
class MyClass {
...
function getData($query) {
$result = mysqli_query($this->conn, $query);
while($row=mysqli_fetch_assoc($result)) {
$resultset[] = $row;
}
if(!empty($resultset))
return $resultset;
} }
and then use the class object to call function in your code
然后使用类对象调用代码中的函数
<?php
$obj = new MyClass();
$row = $obj->getData("select city_name from city");
?>
<select>
<?php foreach($row as $row){ ?>
<option><?php echo $row['city_name'] ?></option>
<?php } ?>
</select>
