Question

我有2张桌子

usrfrnd_table

user_id friend_id
__________________
1         2

2         5

3          1

2          1

和usr_table

id fullname
________________

1  John Doe

2 George Cann
...

如何使用循环从表usr_table回显所有朋友，例如id 2？尝试了这段代码，但它不起作用：

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

试过这个也行不通：

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

我哪里错了？

<?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>

Answer 1

我相信麻烦在于你的PHP，而不是你的SQL。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扩展程序。在这种情况下，$db->query()返回结果对象，而不是数组。您必须在结果对象上重复调用fetch_array()以获取结果集中的每一行。

此外，我建议您将此作为查询使用（但这不是直接问题）：

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;

Answer 2

使用user_id加入表usrfrnd_table，而不是friend_id：

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

Answer 3

摆脱$ id附近的单个撇号，它是一个int，所以它们不是必需的。此外，您需要在$ result上调用fetch_assoc并检查以确保它不是NULL：

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

  $result->free();
}

加入mysql表

3 个答案: