为什么连接了mysql却查询不了,该怎么处理

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发布： 2016-06-13 13:44:34

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为什么连接了mysql却查询不了
表单页代码：

PHP code

<!--

Code highlighting produced by Actipro CodeHighlighter (freeware)
http://www.CodeHighlighter.com/

--><html>
<head>catalog search</head>
<body>
<h1>catalog search</h1>
<form action="results.php" method="post">
choose search type:<br/>
<select name="searchtype">
<option value="author">author</option>
<option value="title">title</option>
<option value="isbn">isbn</option>

</select>
<br/>
enter search term:<br/>
<input name="searchterm" type="text" size="40"/>
<br/>
<input  type="submit" name="submint" value="search"/>
</form>
</body>

</html>

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PHP code

<!--

Code highlighting produced by Actipro CodeHighlighter (freeware)
http://www.CodeHighlighter.com/

--><html>
<body>
<h1>search results</h1>
<?php
$searchtype=$_POST['searchtype'];
$searchterm=trim($_POST['searchterm']);
if(!$searchtype||!$searchterm){
    echo"请输入值";
        exit;

}
if(!get_magic_quotes_gpc())
{
    $searchtype=addslashes($searchtype);
    $searchterm=addslashes($searchterm);
}
@ $db=new mysqli('localhost','root','root','books');
if (mysqli_connect_errno()) 
{
echo 'Error: Could not connect to database. Please try again later.';
exit;
}

$query="select * from books where".$searchtype."like '%".$searchterm."%'";
$result=$db->query($query);
$num_results=$result->num_rows;
echo "<p>number of books found:".$num_results."</p>
                    <div class="aritcle_card">
                        <a class="aritcle_card_img" href="/ai/900">
                            <img src="https://img.php.cn/upload/ai_manual/000/000/000/175679989420709.png" alt="知了zKnown">
                        </a>
                        <div class="aritcle_card_info">
                            <a href="/ai/900">知了zKnown</a>
                            <p>知了zKnown：致力于信息降噪 / 阅读提效的个人知识助手。</p>
                            <div class="">
                                <img src="/static/images/card_xiazai.png" alt="知了zKnown">
                                <span>65</span>
                            </div>
                        </div>
                        <a href="/ai/900" class="aritcle_card_btn">
                            <span>查看详情</span>
                            <img src="/static/images/cardxiayige-3.png" alt="知了zKnown">
                        </a>
                    </div>
                ";
for ($i=0;$i<$num_results;$i++)
{
    $row=$result->fetch_assoc();
    echo"<p><strong>".($i+1)."title:";
    echo htmlspecialchars(stripslashes($row['title']));
    echo "</strong><br/>author:";
    echo stripslashes($row['author']);
    echo"<br/>isbn:";
    echo stripslashes($row['isbn']);
    echo"<br/>price:";
    echo stripslashes($row['price']);
    echo"</p>";
}
?>
</body>
</html>

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输出结果：

search results
Number of books found:

为什么查询不出结果

------解决方案--------------------
你没有在执行查询后进行检错，应补上
1、确认 $searchtype 是正确的字段名
2、$query="select * from books where".$searchtype."like '%".$searchterm."%'";
中 like 前少了个空格
写成这样是不是清爽些？
$query = "select * from books where $searchtype like '%$searchterm%'";

------解决方案--------------------
$query="select * from books where".$searchtype."like '%".$searchterm."%'";
echo $query;就知道是否等价了。红字部分连在一起了还对吗？
------解决方案--------------------
$query="select * from books where".$searchtype."like '%".$searchterm."%'";
$query = "select * from books where $searchtype like '%$searchterm%'";
这两个sql是有区别的。
""里面的都是字符串，然后你没有空格直接写.$searchtype。。。。你自己想想结果一样吗？

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