如何将下拉列表（HTML）中的选定值传递给 PHP

Question

我正在动态构建一个选择选项。我从选择选项中选择一个值。 （列表的值来自 php 如何将选定的值传递给 PHP？

<select id ="s1" name="swimopt" class="so">
    <?php echo $options; ?>
</select>' 

$options 来自 MySQL 并填充下拉列表

当我选择一个值，然后尝试

echo $_POST['swimopt'];

不显示所选值

请帮忙

 <form id="swimdata"  method="POST" action="save.php">

     <input type="radio" style="font-size: 16px;font-weight: bolder" name="gender" class ="gender" value="boys">BOYS
     <input type="radio" style="font-size: 16px;font-weight: bolder" name="gender" class ="gender"  value="girls">GIRLS

     <table id="meetTable" style="width:auto">


    <tr>
        <th>EVENT:</th>
        <th>NAME:</th>
        <th>LANE:</th>
        <th>TIME:</th>
        <th>PLACE:</th>
        <th>SCORE 1:</th>
        <th>SCORE 2:</th>
        <th>PLACE:</th>
        <th>TIME:</th>
        <th>LANE:</th>
        <th>NAME:</th>

    </tr>

</table>

     <input type="submit" name="formSubmit" value="Submit" />
     <input type="hidden" name="action1" value="addSwimmer" id="action1">

 </form>

这是我的 PHP 从 mySQL 获取 $options

if ($result->num_rows > 0) {
    $options= '';
    // output data of each row
    while($row = $result->fetch_assoc()) {
        $options .=  "<tr><td>" . $row["swimmer_id"]. "</td><td>" . $row["first_name"]. " " . $row["last_name"]. " </td><td> "  . $row["school_name"]. "</td></tr>";
    }

} else {
    echo "0 results";
}
echo $options;
$conn->close();

截图：

选项回显到选择框中：

<select id ="s1" name="swimopt" class="so">
    <?php echo $options; ?>
</select>' 

PHP文件代码

$sql = "select  first_name, last_name  from swimming where gender='m'";

$results = mysqli_query($link,$sql);

if ($results->num_rows > 0) {
    $options = '';
    // output data of each row
    while($row = $results->fetch_assoc()) {
        $fname=$row["first_name"];
        $lname=$row["last_name"];

        $options .= "<option value= >" . $row["first_name"] . "  " . $row["last_name"]."</option>";


    }
} else {
    echo "0 results";
}
echo $options;
$link->close();


?>

Answer 1

我想我明白你的问题是什么了。您没有设置选择选项。 试试这个：

<?php
$options = 'John';
?>

<form id="swimdata"  method="POST" action="save.php">

<input type="radio" style="font-size: 16px;font-weight: bolder" name="gender" class ="gender" value="boys">BOYS
<input type="radio" style="font-size: 16px;font-weight: bolder" name="gender" class ="gender"  value="girls">GIRLS

<table id="meetTable" style="width:auto">


<tr>
<th>EVENT:</th>
<th>NAME:</th>
<th>LANE:</th>
<th>TIME:</th>
<th>PLACE:</th>
<th>SCORE 1:</th>
<th>SCORE 2:</th>
<th>PLACE:</th>
<th>TIME:</th>
<th>LANE:</th>
<th>NAME:</th>

</tr>

</table>
<select id ="s1" name="swimopt" class="so" value="">
<?php
for($i=0;$i<10;$i++){
echo '<option value="user '.$i.'" >user '.$i.'</option>';
}
?>
</select> 
<input type="submit" name="formSubmit" value="Submit" />
<input type="hidden" name="action1" value="addSwimmer" id="action1">

</form>

Answer 2

select 选项必须用带有value 属性的html 标签<option> 包裹，如果它被选中将被发布。 根据您给定的代码，您不会为select 生成options。所以尝试这样的事情：

if ($result->num_rows > 0) {
$options= '';
// output data of each row 
while($row = $result->fetch_assoc()) {
    $options .=  "<option value=\"". $row["swimmer_id"] ."\">" . $row["first_name"]. " " . $row["last_name"]. " - "  . $row["school_name"]. "</option>";
}

} else {
echo "0 results";
}
echo $options;
$conn->close();

现在如果你$_POST 它你会得到swimmer id