SQL内部联接问题

本文介绍了SQL内部联接问题的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

$link=mysql_connect('localhost','root','12345') or die (mysql_error());

mysql_select_db("library") or die (mysql_error());



?>

<html>

    <head>

        <meta http-equiv="content-type" content="text/html; charset=ISO-8859-1" />

        <title>AIMST Online Library Systeme</title>



        <link href="/library/css/style.css" rel="stylesheet" type="text/css" />

    </head>



    <body>

        <div id="wrapper">

<?php include_once $_SERVER['DOCUMENT_ROOT'] .'/library/includes/header.php'; ?>

                <div id="page">

                <div id="content">

                    <div id="welcome">



 <h2 class="head2" ><a class="head">View / Edit Book Master Data</a></h2>



 <table class="aatable">

<tr>

<th >SL No</th>

<th>Title</th>

<th>Media Type</th>

<th >Author</th>

<th >Publication</th>

<th >Edition</th>

<th >Year</th>

<th >No of Copy Avail</th>

<th >Location</th>

<th >Category</th>

<th >Sub Category</th>

<th >Loan Status</th>

<th >Remark</th>

</tr>

<?php



if ($_POST[search_type]=='all'){

$query = "SELECT * FROM book WHERE

slno LIKE '%$_POST[keyword]%'||

title LIKE '%$_POST[keyword]%' ||

author LIKE '%$_POST[keyword]%'||

publication LIKE '%$_POST[keyword]%'||

year LIKE '%$_POST[keyword]%'||

location LIKE '%$_POST[keyword]%'||

category LIKE '%$_POST[keyword]%'||

sub_category LIKE '%$_POST[keyword]%'||

related_name LIKE '%$_POST[keyword]%'||

media_type  LIKE '%$_POST[keyword]%'||

book_remark LIKE '%$_POST[keyword]%'

    ";

$result = mysql_query($query,$link)

or die(mysql_error());

}



else {

$query = "SELECT * FROM book WHERE $_POST[search_type] LIKE '%$_POST[keyword]%'";

$result = mysql_query($query,$link)

or die(mysql_error());

}





while ($row = mysql_fetch_array($result)) {



$book_id = $row['book_id'];

$slno = $row['slno'];

$title = $row['title'];

$media_type = $row['media_type'];

$author=$row['author'];

$publication=$row['publication'];

$edition=$row['edition'];

$year=$row['year'];

$total_holding=$row['total_holding'];

$location=$row['location'];

$category=$row['category'];

$sub_category=$row['sub_category'];

$book_remark=$row['book_remark'];



echo "<tr>";

echo "<td>".$slno."</td>";

echo "<td >".$title."</td>";

echo "<td>".$media_type."</td>";

echo "<td>".$author."</td>";

echo "<td>".$publication."</td>";

echo "<td >".$edition."</td>";

echo "<td>".$year."</td>";

echo "<td >".$total_holding."</td>";

echo "<td>".$location."</td>";

echo "<td>".$category."</td>";

echo "<td>".$sub_category."</td>";



echo "<td>".$book_remark."</td>";



echo "</tr>";

}

?>

 </table>

                          <!--body ends-->

 </div>

                    <!-- end div#welcome -->



                </div>

                <!-- end div#content -->



                <!-- end div#sidebar -->

                <div style="clear: both; height: 1px"></div>

            </div>

<?php include_once $_SERVER['DOCUMENT_ROOT'] . '/library/includes/footer.php'; ?>

        </div>

        <!-- end div#wrapper -->

    </body>

</html>

这是我的编码，从一个表名称簿中查询，现在我打算添加一个表调用贷款，并只将上表的coluum调用loan_status加入
我尝试了语法错误..我可以知道吗?

this is my coding, query from a table name book, and now i plan to add in a table call loan and just join the coluum call loan_status with the table above
im tried and got syntax error..may i know how?

coluum

SQL内部联接问题

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