如何使用mysql表中的数据填充选择框 [英] How to populate a selection box using data from an mysql table

查看:214
本文介绍了如何使用mysql表中的数据填充选择框的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我想填充一个下拉列表,其中包含MySQL表中的名称。一切似乎都有效,但名单总是一个名字短。



我尝试过:

<表格代码

I want to populate a dropdown list with the names held in a MySQL table. All seems to work but the list is always one name short.

What I have tried:

Form Code

<p>Select Player from dropdown list: (Required) <select id="FullName"  name="FullName" 
            <?php 
            playerFullName_load()
            ?>
        >
        </select></p>





功能代码:



Function Code:

function playerFullName_load(){
    require '../../configure.php';
        $uName = "";
     $Surname = "";
     $FullName = "";
    $ID = "";
$db_handle = mysqli_connect(DB_SERVER, DB_USER, DB_PASS );
$db_found = mysqli_select_db($db_handle, DB_NAME);
// Check connection
if (!$db_found) {
    die("Connection failed: " . mysqli_connect_error());
}
if ($db_found) {
   
$SQL = "SELECT * FROM `snravalonplayerstb` ORDER BY `FirstName` ASC, `Surname` ASC";
$result = mysqli_query($db_handle, $SQL);
while ( $db_field = mysqli_fetch_assoc($result) ) {
    $uName = $db_field['FirstName'];
    $Surname = $db_field['Surname'];
    $_SESSION['TelephoneNo'] = $db_field['TeleNo'];
    $FullName = $uName." ". $Surname;
    echo "<option value='$FullName'> $FullName </option>";
}
echo "<option disabled selected value> -- select an option -- </option>";
}
else {
    print "Database NOT Found ";
}
mysqli_close($db_handle);
}

推荐答案

uName =;


Surname =;
Surname = "";


FullName =;
FullName = "";


这篇关于如何使用mysql表中的数据填充选择框的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

查看全文
登录 关闭
扫码关注1秒登录
发送“验证码”获取 | 15天全站免登陆