php - 连接mysql表

我有两张桌子
usrfrnd_表

user_id friend_id
__________________
1         2

2         5

3          1

2          1

和usrúu表

id fullname
________________

1  John Doe

2 George Cann
...

如何从带循环的usr_表中回显例如id 2的所有朋友？
尝试了这段代码，但不起作用：

$stmt = "SELECT fullname FROM usr_table INNER JOIN usrfrnd_table ON usr_table.id=usrfrnd_table.friend_id WHERE usrfrnd_table.user_id='$id'";
    $result=$db->query($stmt);
    foreach($result as $rec){
        echo "<li><a href=user.php?id=".$rec['id']."'>". $rec['fullname'] . "</a></li>";
    }

试过这个也不行：

$stmt = "select fullname from usr_table inner join usrfrnd_table ON (usr_table.id = usrfrnd_table.friend_id) WHERE usrfrnd_table.user_id ='$id'";
$result=$db->query($stmt);
foreach($result as $rec){
echo "<li><a href=user.php?id=".$rec['id']."'>". $rec['fullname'] . "</a></li>";

我错在哪里？

<?php
error_reporting(E_ALL);
require "db.php";

if(isset($_GET['id']))
{
    $id=$_GET['id'];
    $result=$db->query("SELECT * FROM usr_table WHERE id='$id' LIMIT 1") or die(printf("Səhv1: %s\n", $db->error));
    $data=$result->fetch_object();
    $fullname=$data->fullname;
    $dob = $data->dob;
    $phone= $data->phone;
    $adress= $data -> adress;
    $school = $data->school;
    $info = $data->info;}

?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title><?=$fullname?> | BAZA</title>
<link type="text/css" href="styles/user.css" rel="stylesheet" />

</head>

<body>
<body>
    <div id="address"><?=$adress?></div>
    <h1><?=$fullname?></h1>
    <div id="contact"><?=phone?><a href="" target="_blank" muse_scanned="true">emailaddress@uwaterloo.ca</a></div>

    <div class="section">
        <h2>Summary of Qualifications</h2>
            <ul>
                <li>Bullet one Bullet one</li>
                <li>Bullet two Bullet two</li>
                <li>Bullet three Bullet three</li>
                <li>Bullet four Bullet four</li>
                <li>Bullet five Bullet Five</li>
            </ul>
    </div>

    <div class="section">
        <h2>Work Experience</h2>
        <div class="content">
            <span class="date">Start » End Year</span>
            <h3>Job Title One</h3>
            <address>Company Name, Address, City Province</address>
            <ul>
                <li>Bullet one Bullet one</li>
                <li>Bullet two Bullet two</li>
                <li>Bullet three Bullet three</li>
                <li>Bullet four Bullet four</li>
            </ul>

            <span class="date">Start » End Year</span>
            <h3>Job Title Two</h3>
            <address>Company Name, Address, City Province</address>
            <ul>
                <li>Bullet one Bullet one</li>
                <li>Bullet two Bullet two</li>
                <li>Bullet three Bullet three</li>
                <li>Bullet four Bullet four</li>
            </ul>

            <span class="date">Start » End Year</span>
            <h3>Job Title Three</h3>
            <address>Company Name, Address, City Province</address>
        </div>
    </div>

    <div class="section">
        <h2>Yaxın adamlar</h2>
        <div class="content">
            <span class="date">Dostlar</span>
            <h3>Volunteer Job Title</h3>
            <address>Company Name, Address, City Province</address>
            <ul>
            <?php
$stmt = "select U.* from usr_table U
join usrfrnd_table F on U.id=F.friend_id
where F.user_id='$id'
union distinct select U.* from usr_table U
join usrfrnd_table F on U.id=F.user_id
where F.friend_id='$id'";
    $result=$db->query($stmt);
    foreach($result as $rec){
        echo "<li><a href=user.php?id=".$rec['id']."'>". $rec['fullname'] . "</a></li>";
    }

                ?>
            </ul>
        </div>
    </div>

    <div class="section">
        <h2>Təhsil</h2>
            <ul>
                <li>Candidate for Bachelor of Major, University, Province,  StartYear - Present</li>
                <li>OSSD, Highschool, Town, Province, StartYear-EndYear</li>
            </ul>
    </div>

    <div class="section">
        <h2>Academic Awards</h2>
            <ul>
                <li>Year, Year, Name of award/Scholarship</li>
            </ul>
    </div>
</body>
</html>

最佳答案

我相信问题出在你的PHP上，而不是SQL上。试试这个：

$result = $db->query($stmt);
while ($rec = $result->fetch_array()) {
    echo "<li><a href=user.php?id=".$rec['id']."'>". $rec['fullname'] . "</a></li>";
}

您似乎正在使用mysqli扩展。在这种情况下，$db->query()返回结果对象，而不是数组。必须反复调用结果对象上的fetch_array()，才能获取结果集中的每一行。
另外，我建议您使用此作为查询（但这不是当前的问题）：

select U.* from usr_table U
join usrfrnd_table F on U.id=F.friend_id
where F.user_id=2
union distinct select U.* from usr_table U
join usrfrnd_table F on U.id=F.user_id
where F.friend_id=2;

关于php - 连接mysql表，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/6399129/