如何使用Index.php上的链接显示我的Search.php页面的结果?
<a href="search.php?country_select=spain">Spain</a>
以上操作无效,所有国家/地区都显示在Search.php中。对不起,我是PHP和MySQL的新手。
我的链接在index.php上,是搜索并显示在search.php上。我在search.php上有以下代码:
<?php $appmt = $bsiCore->ClearInput(base64_decode($_REQUEST['appmt_id']));
$apartDetails = $bsiCore->getApartmentdetails($rowappt['appmt_id']);
$featuresarr = $bsiCore->getApmtFeatures($rowappt['appmt_id']);
?>
<div class="col-md-4 col-sm-6 single listing">
<div class="offer offer-wrap">
<img class="img-responsive" src="<?php echo ($rowappt['default_img']=="")? "http://www.placehold.it/800x450/ddd/bbb&text=Image":"gallery/ApartImage/".$rowappt['default_img']; ?>" alt="">
<span class="label label-info">
<?php echo $rowappt['country']; ?>
</span>
<div class="btn-group">
<a href="<?php echo str_replace(" ","-",strtolower(trim($rowappt['appmt_name'])))."-".$rowappt['appmt_id'].".html"; ?>"><button class="btn btn-success btn-sm" type="button">Details</button>
<button class="btn btn-default btn-sm" type="button"><em class="icon-info-sign"></em></button></a>
</div>
<div class="padding img1">
<h2 class="text-center text-info"><?php echo $rowappt['city']; ?></h2>
<h4 class="text-center"><?php echo $rowappt['bedroom']; ?> bedroom, <?php echo $rowappt['bathroom']; ?> bathroom</h4>
<span class="label2 label-info">
<?php echo $bsiCore->config['conf_currency_symbol'].$rowappt['price']; ?><span>p/w*
</span>
</div>
</div>
</div>
表格如下:
<form action="search.php" class="form-inline reservation-horizontal clearfix" role="form" method="post" name="form1" id="form1" style="padding-bottom:0px">
<div class="row">
<div class="col-lg-12" style="min-height: 60px; background-color: rgba(255, 255, 255, 0.4); border-radius: 5px;">
<div class="pad1 col-md-2 col-sm-3 col-xs-12">
<select name="country_select" id="drop_1">
<option value="" selected="selected">All Countries</option>
</select>
</div>
<div>
<div class="pad1 col-md-2 col-sm-3 col-xs-12">
<select name="country_select">
<option value="" selected="selected" disabled="disabled">Any City / Town</option>
</select> </div></div>
<div class="pad1 col-md-2 col-sm-3 col-xs-12">
<select id="appartment_type" name="appartment_type">
<?php echo $bsiCore->getApmtTypeCombobox();?>
</select>
</div>
<div class="pad1 col-md-2 col-sm-3 col-xs-12">
<select name="bathroom" id="bathroom" >
<option value="">Sleeps</option>
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
<option value="5">5</option>
<option value="6">6</option>
<option value="7">7</option>
<option value="8">8</option>
<option value="9">9</option>
<option value="10">10</option>
<option value="11">11</option>
<option value="12">12</option>
<option value="13">13</option>
<option value="14">14</option>
<option value="15">15</option>
<option value="See Description">16+</option>
</select>
</div>
<div class="pad1 col-md-2 col-sm-3 col-xs-12">
<button value="Search" class="btn btn-success" style="width: 100%;">Search</button>
</div>
</div>
</div>
</div>
</form>
答案 0 :(得分:1)
我开始写一个SSCCE来回答你的问题,但我意识到我根本没有足够的信息,但仍然不太明白你在问什么。
您是否在index.php
上有一个HTML表单,其中的操作会将其提交给search.php
?或者,您是否有用户可以点击的链接或一组链接,并且所选的链接控制用户在发送到search.php
时看到的内容?
假设您在index.php
上有一组向GET
页面发送search.php
请求的链接(如您的初始代码行?country_select=spain
所示),然后你使用这个特定的请求来查询数据库?或者您是使用存储在search.php
页面上的变量/数组中的一组数据吗?
如果您要提交表单,它应该类似于:
index.php
<form method="post" action="search.php">
<label>Select Country:</label><br>
<input type="text" id="country"><br>
<input type="submit" value="Submit">
</form>
search.php
<?php
$country=$_POST['country'];
$con = mysqli_connect("localhost","my_user","my_password","my_db");
$query = "SELECT * FROM `country_table` WHERE `country_name`= '".$country."'";
mysqli_query($con, $query);
?>
如果您使用多个链接选项,它可能如下所示:
index.php
<p>Please select a country</p><br>
<a href="search.php?country_select=spain">Spain</a><br>
<a href="search.php?country_select=united+states">United States</a><br>
<a href="search.php?country_select=italy">Italy</a><br>
search.php
<?php
$country=$_GET['country_select'];
$con = mysqli_connect("localhost","my_user","my_password","my_db");
$query = "SELECT * FROM `country_table` WHERE `country_name`= '".$country."'";
mysqli_query($con,$query);
?>
同样,我不知道您的数据是如何存储/格式化的,我不知道您希望如何向用户呈现搜索,我不知道您希望如何处理结果。如需更准确的回复,请详细说明问题和预期结果。
答案 1 :(得分:0)
确保&#34; country_select&#34;在您的PHP接收代码中为&#34; $ _ REQUEST [&#39; country_select&#39;];&#34;。否则,它可能是search.php文件的解析端的问题。处理MySQL查询时也可能是一个问题。