Question

我还在学习功能，而且有点不确定。

我这里有这个代码：

<?php
include ('blah.html');
$conn = mysqli_connect("localhost","root","","testing");
// Check connection
if (mysqli_connect_errno())
{
    echo "<p>Failed to connect to MySQL: " . mysqli_connect_error() . "</p>";
}
else
{
    echo "<p>Connected to mysql and the testing database</p>";
}

/*function produceTable(){
    $result = mysqli_query($conn,"SELECT studentid, CONCAT(firstname,' ', lastname) as name,phone, email, courseid FROM students WHERE courseid > 1500 ORDER BY lastname");
    return $result;
}*/
$result = mysqli_query($conn,"SELECT studentid, CONCAT(firstname,' ', lastname) as name,phone, email, courseid FROM students WHERE courseid > 1500 ORDER BY lastname");
$numRecords = mysqli_num_rows($result);
while($row = mysqli_fetch_array($result)) {
    echo "<tr>";
    echo "<td align = 'center'>".$row['studentid']."</td>";
    echo "<td>".$row['name']."</td>";
    echo "<td>".$row['phone']."</td>";
    echo "<td>".$row['email']."</td>";
    echo "<td>".$row['courseid']."</td>";
    echo "</tr>";
}
echo "</table>";
echo "<p>".$numRecords." records found with course id larger than 1500</p>";                    
?>

</body>
</html>

正如您所看到的，我已经注释掉了这个功能。如果忽略该代码的功能并输出我想要的记录，但我需要能够使它与函数一起工作。该函数需要被称为produceTable（）并接受一个名为$ result的参数。它还需要使用mysqli_fetch_array（）来输出记录。

Answer 1

你的功能是下一个：

function produceTable(){
    $result = mysqli_query($conn, "SELECT ...");
    return $result;
}

$conn没有任何价值。不是吗？尝试将此参数传递给函数

Answer 2

在produceTable() $conn中undefined。只需将$conn作为参数传递即可。像这样

function produceTable($conn){
    $result = mysqli_query($conn,"SELECT studentid, CONCAT(firstname,' ', lastname) as name,phone, email, courseid FROM students WHERE courseid > 1500 ORDER BY lastname");
    return $result;
}

$result = produceTable($conn);
$numRecords = mysqli_num_rows($result);
while($row = mysqli_fetch_array($result)) {
    .....
}

函数使用mysqli_fetch_array输出数据库记录

2 个答案: