php 从 MySQL 数据库中获取数据到 html 下拉列表
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Fetching data from MySQL database to html drop-down list
提问by Meet Bhatt
I search for answers here but haven't found a solution. I have also added picture of the error. I want the data to go to the first drop-down list ( above the error) I think the method I try to perform is also create drop-down list, am I correct?
我在这里搜索答案，但还没有找到解决方案。我还添加了错误的图片。我希望数据转到第一个下拉列表（错误上方）我认为我尝试执行的方法也是创建下拉列表，对吗？
<form name="message" action="" method="post" onsubmit="" accept-charset="utf-8">
<div class="form-group">
<label id="senderName">?? ?????:</label>
</div>
<div class="form-group">
<label for="to_user">???:</label>
<select name="to_user" class="form-control">
<option value="pick">??? ???????</option>
<?php
$sql = \mysqli_query("SELECT name From users");
$row = mysqli_num_rows($sql);
echo "<select name='to_user'>";
while ($row = mysqli_fetch_array($sql)){
echo "<option value='". $row['name'] ."'>" .$row['name'] ."</option>" ;
}
echo "</select>" ;
?>
</select>
</div>
回答by The Codesee
In MySQLi, the first parameter of a query needs to be the database connection. Also, there's no need to add a \before the statement.
在 MySQLi 中，查询的第一个参数需要是数据库连接。此外，无需\在语句前添加 a 。
$sql = \mysqli_query("SELECT name From users");should be $sql = mysqli_query($con, "SELECT name From users");
$sql = \mysqli_query("SELECT name From users");应该 $sql = mysqli_query($con, "SELECT name From users");
Note: replace $con with your database connection variable!
注意：将 $con 替换为您的数据库连接变量！
As you mentioned that you wanted the result from the database to go inside the selectform, simply adjust your code to look like this:
正如您提到的，您希望将数据库中的结果放入select表单中，只需将代码调整为如下所示：
<select name="to_user" class="form-control">
<option value="pick">??? ???????</option>
<?php
$sql = mysqli_query($con, "SELECT name From users");
$row = mysqli_num_rows($sql);
while ($row = mysqli_fetch_array($sql)){
echo "<option value='". $row['name'] ."'>" .$row['name'] ."</option>" ;
}
?>
</select>
回答by Raj Singh
<div class="row form-group">
<div class="col col-md-3">
<label for="email-input" class=" form-control-label"> Vehicle</label>
</div>
<div class="col-12 col-md-9">
<select name="car_id" id="car_id" class="form-control-label" >
<?php
$list = mysqli_query($conn,"SELECT * FROM `vehicle_registration` where `status`='0' ");
while ($row_ah = mysqli_fetch_assoc($list)) {
?>
<option value="<?php echo $row_ah['id']; ?>"><?php echo $row_ah['car_no']; ?></option>
<?php } ?>
</select>
</div>
</div>
回答by Meet Bhatt
<label><b>Select Steam: </b></label>
<select id="study">
<option value="" selected="selected" disabled="">---Selected---</option>
<?php
$query = "SELECT study FROM details";
$query_run = mysqli_query($con, $query);
while ($row = mysqli_fetch_array($query_run)) {
echo "<option value='".$row['study']."'>".$row['study']."</option>";
}
?>
</select>
