php下拉列表未选择值

vql8enpb 于 2021-06-21 发布在 Mysql

关注(0)|答案(3)|浏览(471)

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两年前关门了。
我使用php从数据库中检索值，但遇到了一个问题。当我从下拉列表中选择一个值并按submit时，我输入的值不会被选中。表单中的所有其他元素都能完美地工作。我不确定我做错了什么。
当使用isset时，最终结果显示为nothing，而不是我在下拉菜单中选择的内容。
这是我的密码：

<?php 
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "timedrun";
$conn = new mysqli($servername, $username, $password, $dbname);

if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} else {
echo "Connected successfully";
}
?>

<html>

<head>

<title>Add Run</title>

<link href='https://fonts.googleapis.com/css?family=Roboto' rel='stylesheet'> 
<link rel='stylesheet' type='text/css' href='mystyle.css'>

</head>

<body>
<h1>Timed Run</h1>

<center>
<hr>

<form action='#' method='post'>
Select Student: <select name='studentSelected' class="form">

<?php
$fsql = "SELECT firstName FROM students";
$fresult = $conn->query($fsql);
$lsql = "SELECT lastName FROM students";
$lresult = $conn->query($lsql);

if ($fresult->num_rows > 0) {
    while ($row = $fresult->fetch_assoc()){
        echo '<option value="'.$row['ID'].'">' .$row['firstName'].' </option>';
    }
};

?>
</select>

<br><a class="space" ></a><br>

Time (s): <input type="int" name="time" class="form">

<br><a class="space"></a><br>

Select Date: <input type="date" name="date" class="form">

<br><a class="space"></a><br>

Notes: <input type="text" name="notes" class="form">

<br><a class="space"></a><br>

<input type="submit" class="form" name="submit">

</form>

<?php
if(isset($_POST['submit'])){
$selected_val = $_POST['studentSelected'];
echo "You have selected :" .$selected_val;
}
?>
</center>
</body>
</html>

mysql php Html html-select

来源：https://stackoverflow.com/questions/48113651/php-dropdown-list-not-selecting-value

3条答案

按热度按时间

yfjy0ee71#

<?php
$fsql = "SELECT firstName FROM students";
$fresult = $conn->query($fsql);
$lsql = "SELECT lastName FROM students";
$lresult = $conn->query($lsql);

if ($fresult->num_rows > 0) {
    while ($row = $fresult->fetch_assoc()){
        echo '<option value="'.$row['ID'].'">' .$row['firstName'].' </option>';
    }
};

您的查询只用于从表中选择firstname，并使用 value="'.$row['ID'].'" . 将查询更改为

SELECT ID,firstName FROM students

赞(0）回复(0）举报 2021-06-21

sr4lhrrt2#

这是因为你没有选择 ID 在您的查询中。将查询更改为：

$fsql = "SELECT ID,firstName FROM students"; // add ID
$fresult = $conn->query($fsql);
$lsql = "SELECT ID,lastName FROM students";// add ID
$lresult = $conn->query($lsql);

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vktxenjb3#

现在应该可以了

<?php
// $fsql = "SELECT * FROM students";
$fsql = "SELECT ID,firstname FROM students";
$fresult = $conn->query($fsql);

if ($fresult->num_rows > 0) {
    while ($row = $fresult->fetch_assoc()){
        echo '<option value="'.$row['ID'].'">' .$row['firstName'].' </option>';
    }
};

赞(0）回复(0）举报 2021-06-21

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php下拉列表未选择值

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