Question

我试图找到插入表中的一组成绩的gpa，成绩的列是varchar类型，我需要循环然后将它们转换为十进制，我很丢失，现在不怎么去对这个。任何帮助将不胜感激

继续我到目前为止，我一直在犯错：

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" 

"http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
    <head>
        <title>Display GPA</title>
    </head>
    <body>
<?php

$host="XXXX"; 
$username="XXXX"; // Mysql username 
$password="XXX"; // Mysql password 
$db_name="XXXX"; // Database name 
$tbl_name="gpa"; // Table name

// Connect to server and select databse.
$con=mysqli_connect("$host", "$username", "$password")or die("cannot connect");
mysqli_select_db($con,"bparis") or die ("no database");

if(mysqli_connect_errno())
$con = false;

if(!$con) {
    echo "Failed to connect to database server!";

}

else {




    $result = mysqli_query($con,"SELECT grade FROM gpa");
while ($row = mysql_fetch_array($result, MYSQL_NUM)) {
    if($grade == 'A') {
        $numberGrade = 4;
    } elseif ($grade == 'A-') {
        $numberGrade = 3.67;
    } elseif ($grade == 'B+') {
        $numberGrade = 3.33;    
} elseif ($grade == 'B') {
    $numberGrade = 3;    
} elseif ($grade == 'B-') {
    $numberGrade = 2.67;
} elseif ($grade == 'C+') {
    $numberGrade = 2.33;                                                                                       
} elseif ($grade == 'C')  {
    $numberGrade = 2;
} elseif ($grade == 'C-') {
    $numberGrade = 1.67;                                                                                   
} elseif ($grade == 'D')  {
    $numberGrade = 1;                                                                           
} else {
    $numberGrade = 0;                                                                           
}
$totalGrade = ($totalGrade + $numberGrade)/$i;
array_pop($array);
}
$totalGrade;

    echo "        <h1>Your Grades and GPA Calculator</h1>
    <hr />
    <p />
    <table border='1'>
        <tr>
            <th></th>
            <th>Course Heading</th>
            <th>Course Number</th>
            <th>Credit Hours</th>
            <th>Grade</th>
        </tr>\r\n";
       $result = mysqli_query($con, "SELECT * FROM gpa");
    while($row = mysqli_fetch_array($result)) {
        echo "            <tr>\r\n";
        echo "                <td>" . $row['id'] . "</td>\r\n";
        echo "                <td>" . $row['heading'] . "</td>\r\n";
        echo "                <td>" . $row['courseNUM'] . "</td>\r\n";
        echo "                <td>" . $row['creditHours'] . "</td>\r\n";
        echo "                <td>" . $row['grade'] . "</td>\r\n";
        echo "            </tr>\r\n";
    }
    echo "        </table>\r\n"; 
    echo "<p />Your overall GPA:" .$totalGrade;
}
mysqli_close($con);

?>


</body>

Answer 1

你真的不需要用PHP来做。 SQL非常有能力自行完成

SELECT AVG(
    CASE gpa 
        WHEN 'A' THEN 4
        WHEN 'A-' THEN 3.67
        WHEN 'B+' THEN 3.33
-- .... and so on
        ELSE 0
    END
    ) AS avg_gpa
 FROM table

用php循环浏览mysql表，找到字母等级的平均成绩

1 个答案: