2016-10-01 91 views
0

我有这种形式如何提交表单到同一页面,并得到结果显示

<form action="" method="GET" id="formSearch"> 
      <div class="input-group"> 
       <input type="text" class="form-control" id="inputSearch" name="inputSearch"/> 
       <span id="spanBtnSearch" class="input-group-btn"> 
        <input type="submit" class="btn btn-primary" id="btnSearch" value="Search"/> 
       </span> 
      </div>   
     </form> 

我要发送这种形式在同一个页面,我的连接和查询上低于

<?php 
$conn = mysqli_connect("localhost:3311","root","root","compared"); 
$shingpoint = "SELECT * FROM products WHERE store LIKE 'shingpoint' AND title LIKE '%".$_POST['inputSearch']."%'"; 
$result = mysqli_query($conn,$shingpoint); 
?> 

而且应该显示结果的结果是这样的

<div class="row" id="divShingpoint"> 
     <h2 class="section-title style2 text-left">Shingpoint</h2> 
     <div id="divProductThumbnail" class="item col-lg-4 col-md-4 col-sm-4 col-xs-6 col-xxs-12 colums"> 
      <div class="container" id="divInnerContainer"> 
      <?php while ($shingpointrow = mysqli_fetch_assoc($result)) {?> 
       <div class="product"> 
        <div class="image"><img class="img-responsive" src="<?php echo $shingpointrow['img'] ?>" alt="<?php echo $shingpointrow['title']?>"/></div> 
        <div class="description"><h4 class="productname"><?php echo substr(trim($shingpointrow['title']),0,20) ?></h4></div> 
        <div class="price"> 
         <span><?php echo $shingpointrow['price'] ?></span> 
         <input type="button" class="btn btn-primary btn-sm" value="Details"/> 
        </div> 
       </div> 
      <?php }?> 
      </div> 
     </div> 
    </div>  

它确实发送数据,因为页面刷新并且url已启动过期但结果不显示在我想要搜索的标题div。用C inputSearch:

而且我收到此错误

注意:未定义指数\ wamp64 \第3行

什么缺什么我做错了WWW \ compareit.php?或者有任何其他方法来完成任务?

回答

0

您将方法设置为GET,但您尝试从$ _POST读取。

因此,要么将<form method="POST">$_POST['inputSearch']更换为$_GET['inputSearch']

除此之外,我想补充一些验证,使用PDO和改变一些事情......

0

表单应具有方法而不是得到

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