Question

我正在建立一个日托学生注册系统。我有一个表格，可以将数据提交到我的日托表中。我有一个名为Grade的专栏，我想根据所选的成绩检索所有学生。任何人都可以协助我构建适当的逻辑来做到这一点吗？感谢

这是我的HTML代码

<form method="POST" action="display.php">
View Student Records
<select name="grades">
<option value="all">All</option>
<option value="3">3rd Grade</option>
<option value="4">4th Grade</option>
<option value="5">5th Grade</option>
<option value="6">6th Grade</option>
</select>
<input type="submit" value="View Students" name="view">
</form>

PHP代码如果用户选择三年级学生，这是我尝试的逻辑。我的成绩字段也是文本格式而不是数字

<?php
session_start();
require("dbconnect.php");
if(empty($_SESSION['user_name']))
{
echo "Please login: <a href='./index.php>Login</a>";
exit();
}
?>

<?php 
$link=Connect();
if (isset($_POST['view'])){

$grade = ($_POST['grades']);
$select = "SELECT * daycare WHERE Grade = $grade";
$result = mysql_connect($select, $link) or die("Error: "mysql_error()); 
}


echo "<table>";

while($row = mysql_fetch_array($select){   
echo "<tr><td>" . $row['fname'] . "</td><td>" . $row['lname'] . "</td></tr>" . "</td><td>" . $row['grade'] . "</td></tr>";  
}

echo "</table>"; 
}


mysql_close(); 

?>

Answer 1

hellow也许你可以试试这个

连接数据库的文件：

<?php
function Connect()
{
    if (!($link=mysql_connect("localhost","root","")))
    {
        echo "Error to connect to DataBase.";
        exit();
    }
    if (!mysql_select_db("DataBase_Name",$link))
    {
        echo "Error to select DataBase.";
        exit();
    }
    return $link;
}

?>

文件选择成绩并显示

<form method="post" action="display.php">
View Student Records
<select name="grades">
<option name="all" value="all">All</option>
<option name="third" value="third">3rd Grade</option>
<option name="fourth" value="fourth">4th Grade</option>
<option name="fifth" value="fifth">5th Grade</option>
<option name="sixth" value="sixth">6th Grade</option>
</select>
<input type="submit" value="View" name="view">
</form>

文件选择并返回表格等级：

<?php
session_start();
require("dbconnect.php");
if(empty($_SESSION['user_name']))
     {
           echo "session not started please login: <a href='./login.php'>Login</a>";
             exit();
         }

?>



<?php
$link=Connect(); 

if (isset($POST['view'])){

$query3 = "SELECT * FROM daycare WHERE Grade = "3"" ;
$result3=mysql_query($query3,$link) or die("Error: ".mysql_error());


}


echo "<table>";

while($row = mysql_fetch_array($result3){   
echo "<tr><td>" . $row['fname'] . "</td><td>" . $row['lname'] . "</td></tr>" . "</td><td>" . $row['grade'] . "</td></tr>";  
}

echo "</table>"; 

mysql_close($link); 

?>

你应该用if来验证是否存在值（全部 - 第三 - 第四......）并根据您的选择显示表格等级。

基于用户选择检索mysql表数据并在html表中显示

1 个答案: