问题描述
您好,我试图进行查询,将要进入我在mysql数据库中的4个表中,它将执行一个下拉列表,其中是一个搜索框,然后人们可以单击一个名称,然后它将打开他们搜索的文档.
这是一个问题,即时通讯仅用一个表就能做到这一点,一旦我尝试包括另一个表就行不通了...
我在网上搜索,但看到一些联接查询,但是这些查询不起作用...这是我的1个表的代码正常工作的原因.
这是一个从html页面调用的文件:(livesearch.mysql.php)
Hi im trying to make a query thats gonna go into 4 tables i have in my mysql data base and it will do a drop down list, wich is a search box, then people can click on a name and it will open the document they searched for.
heres the problem, im capable of doing this with only 1 table, as soon i try to include another table it doesnt work...
i searched the web and i saw some join query but those doesnt work... heres my code for 1 table wich is working.
this is a file that is called from a html page :(livesearch.mysql.php)
<?php <br mode="hold" /?> //Get the q parameter from URL
$q=$_GET["q"];
$hint="";
$link = mysql_connect('*server*', '*username*', '*pswd*');
if (!$link)
{
die('Could not connect: '.mysql_error());
}
mysql_select_db("*database*");
$query='SELECT * FROM procedures WHERE Nom LIKE "%'.$q.'%"';
$result = mysql_query($query);
$row_count=mysql_num_rows($result);
while ($row = mysql_fetch_array($result))
{
$appName = $row['Nom'];
$appLink = $row['Lien_Nom'];
if ($hint=="")
{
$hint="<a href="".$appLink."" target="_blank">".$appName."</a>";
}
else
{
$hint=$hint."<br /><a href="".$appLink."" target="_blank">".$appName."</a>";
}
}
if ($hint=="")
{
$response="no suggestions...";
}
else
{
$response=$hint;
}
echo $response;
?>
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