我有四个表j_music，j_topics，J_users，j_video，带有MySQL数据库结构：
J_music
id artist title Views Album art Link uploader
1 niyi Love me 121 Love.jpg Niyi-love.mp3 Admin
2 tolu I am an ode 111 Toludope.jpg Tolu-doe.mp3 admin
3 justin Kind love 122 Justin-kind-love.jpg Justin-kind-lovw.mp3 Admin
J_video
Id artist title Downloads video art Link uploader
1 vandeross Sweet love 121 Love.jpg Niyi-love.mp3 Admin
2 richie Locked down 111 Toludope.jpg Tolu-doe.mp3 Admin2
3 justin A wonderful word 122 Justin-kind-love.jpg Justin-kind-lovw.mp3 Admin
J_user
id username First name Last name userdp password country
1 niyi faleye niyi Niyi-dp.jpg ***** Wales
2 tolu abolarin tolu Tolu-dp.jpg ******* Nigeria
3 justin gayle justin Justin-dp.jpg ******** England
J_topics
id title subject Views Topic image Date creator
1 Justin beiber goes nude ************** 121 Love.jpg 21-02-2013 mark
2 High crisis in the ******** 111 Toludope.jpg 21-02-2013 peter
3 Justin gatlin wins ******** 122 Justin-kind-love.jpg 21-02-2013 Steven
我想实现一个php和mysql搜索，在四个表中搜索特定查询“justin”，如果找到则输出结果。它应该与搜索查询匹配以下列; j_music中的艺术家和标题，j_video中的艺术家和艺术家和标题，j_users中的用户名，j_topics中的标题和主题：
（a）如果在音乐表中找到结果，则提供特殊链接 （b）如果在主题表中找到结果，则它提供特殊链接 （c）如果在视频表中找到结果，则提供特殊链接 （d）如果在users表中找到结果，则它提供一个特殊链接
结果集中有这样的东西
If found in music
$artist $title <a href=’music.php?id’>click here></a>”;
If found in video
$artist $title <a href=’video.php?id’>click here></a>”;
If found in topics
$title $subject <a href=’topic.php?topicid’>click here></a>”;
If found in users
$username $firstname <a href=’users.php?userid’>click here></a>”;
这是我用于单个表搜索和输出的代码：
echo"<form action='#' method='POST'><ul><li><input type='text' name='q' /><input type='submit' name='submit' value='Search' class='button'></center></li></ul>";
$q=$_REQUEST["q"];
if(isset($q) && !empty($q))
{
$q=trim($q);
$q=cleanvalues($q);
$q=strtoupper($q);
//echo"you searched for $q";
//$u="is";
$self=$_SERVER["PHP_SELF"];
$rowsperpage=2;
$range=2;
if(isset($_GET["currentpage"]) && is_numeric($_GET["currentpage"]))
{
$currentpage=(int)$_GET["currentpage"];
}
else
{
$currentpage=1;
}
$offset=($currentpage-1)*$rowsperpage;
$numrows=mysql_num_rows(mysql_query("SELECT * from j_users WHERE UPPER(username) LIKE '%$q%' OR UPPER(username) LIKE '%$q' OR UPPER(username) LIKE '$q%'"));
$totalpages=ceil($numrows/$rowsperpage);
if($currentpage>$totalpages)
{
$currentpage=$totalpages;
}
if($currentpage<1)
{
$currentpage=1;
}
$query=mysql_query("SELECT * from j_users WHERE UPPER(username) LIKE '%$q%' OR UPPER(username) LIKE '%$q' OR UPPER(username) LIKE '$q%' LIMIT $offset, $rowsperpage");
$num=mysql_num_rows($query);
if($num==0)
{
echo"No result found";
}
else
{
echo"Search Results For $q";
while($info=mysql_fetch_assoc($query))
{
$username=cleanvalues2($info["username"]);
$id=$info["userID"];
echo"<a href='profile.php?uid=$id'>$username</a>";
}
if($currentpage>1)
{
echo"<a href='$self?currentpage=1&q=$q'>First </a>";
$prevpage=$currentpage-1;
echo"<a href='$self?currentpage=$prevpage&q=$q'>Prev</a>";
}
for($x=($currentpage-$range); $x<(($currentpage+$range)+1); $x++)
{
if(($x>0) &&($x<=$totalpages))
{
if($x==$currentpage)
{
echo"[<font color='red'>$x</font>]";
}
else
{
echo"<a href='$self?currentpage=$x&q=$q'>[<b>$x</b>]</a>";
}
}
}
if($currentpage!=$totalpages)
{
$nextpage=$currentpage+1;
echo"<a href='$self?currentpage=$nextpage&q=$q'>Next</a>";
echo"<a href='$self?currentpage=$totalpages&q=$q'>Last</a>";
}}
}
答案 0 :(得分：0)
伙计，你的代码真的不好。我建议你选择一个PHP框架并开始研究它，这将对你有所帮助。 Laravel 是我的建议。它易于学习且功能强大。
正如其他用户建议的那样，更喜欢使用 PDO 而不是 mysql _ * 功能。
如果你想做一些 SELECT ，这是一个使用 PDO MySQL 数据库的例子：
$pdo = new PDO('mysql:host=localhost;dbname=mydb', 'myuser', 'mypassword');
$artist = "Justin";
$query = $pdo->query("SELECT id, artist, title FROM j_music WHERE artist LIKE '%".$artist."%'");
$row = $query->fetch(PDO::FETCH_ASSOC);
echo($row['artist']." ".$row['title']." <a href='music.php?'".$row['id'].">click here></a>");