Question

我正在开发一个项目，我需要从html下拉列表中选择选项，然后显示从数据库中选择的数据。就像过滤并显示来自数据库的过滤数据一样。还是行不通我得到错误：

它只是显示来自db的所有数据，而不是通过过滤查询

<form action="search.php" method="POST">
<div class="md-form">
<select class="mdb-select" name="state_search" required>
<option disabled selected>Choose your State</option>
<option name="state" value="Abia State">Abia State</option>
<option name="state" value="Adamawa State">Adamawa State</option>
<option name="state" value="Anambra State">Anambra State</option>
</select>
</div>
<div class="col-lg-4">
<div class="md-form">
<select class="mdb-select" name="school_search">
<option value="" disabled selected>Choose your School</option>
<option value="Michael Opkara University">Mouau</option>
<option value="University of Benin">UniBen</option>
<option value="University of Porthacourt">UniPort</option>
</select>
</div>
<button type="submit" class="btn purple-gradient btn-lg" name="filter">Search <i class="fa fa-paper-plane-o ml-1"></i></button>
</div>
</form>

PHP

        <?php
$school_search = $connect->real_escape_string($_POST["school_search"]);
$state_search = $connect->real_escape_string($_POST["state_search"]);

$query = mysqli_query($connect, "SELECT * FROM roomate WHERE (`state` LIKE '%".$state_search."%') OR (`school` LIKE '%".$school_search."%')") or die(mysql_error());
if($query){
  while($row = mysqli_fetch_array($query)){
    ?>
    <div class="container" style="padding-top:70px;">
    <!--Grid row-->
      <div class="row">
        <?php
        foreach ($query  as $user) {?>
          <div class="col-lg-4 col-md-12 mb-r">
            <!--Featured image-->
            <div class="overlay hm-white-slight z-depth-1-half mb-2">
              <img src="<?php echo $user['room_avatar']?>" class="img-fluid" alt="avatar image">
              <a><div class="mask"></div></a>
            </div>
            <!--Excerpt-->
            <a href="" class="pink-text">
              <h6>
                <i class="fa fa-map"></i><strong><?php echo $user['city'];?>, <?php echo $user['state'];?></strong><br>
                <small><?php echo $user['school'];?></small>
              </h6>
            </a>
            <h4><?php echo $user['type'];?></h4>
            <p>
              by <a><strong><?php echo $user['room_name'];?></strong></a>
              <?php echo $user['created'];?>
            </p>
            <label class="badge badge-danger"> non-verified </label>
            <a href="contact.php">
              <label class="badge badge-primary">Report Scam</label>
            </a>
            <p><?php echo $user['discription'];?></p>
            <a class="btn btn-pink btn-rounded">Contact <?php echo $user['room_name'];?></a>
          </div>
        <?php
        } ?>  
      </div>      
    </div>
    <?php
  }
}
else{
  echo(mysql_error());
}

?>

需要帮助才能完成这项工作

Answer 1

select的名称需要与$_POST变量键匹配。

因此，在您的情况下，<select>需要名称state_search。

或将您的$_POST搜索更改为$_POST['state']以匹配您的选择名称。

您还应该在查询中使用$state_search变量而不是$query。

您可能会遇到更多问题，只需注意细节。

Answer 2

你在echo函数中添加了很多HTML。

<?php
$school_search = $connect->real_escape_string($_POST["school_search"]);
$state_search = $connect->real_escape_string($_POST["state_search"]);

$query = mysqli_query($connect, "SELECT * FROM roomate WHERE (`room_name` LIKE '%".$query."%') OR (`discription` LIKE '%".$query."%') OR (`discription` LIKE '%".$query."%') OR (`type` LIKE '%".$query."%')") or die(mysql_error());;
$result = mysqli_query($connect, $query);
if($result){
  while($row = mysqli_fetch_array($result)){
    ?>
    <div class="container" style="padding-top:70px;">
    <!--Grid row-->
      <div class="row">
        <?php
        foreach ($query  as $user) {?>
          <div class="col-lg-4 col-md-12 mb-r">
            <!--Featured image-->
            <div class="overlay hm-white-slight z-depth-1-half mb-2">
              <img src="<?php echo $user['room_avatar']?>" class="img-fluid" alt="avatar image">
              <a><div class="mask"></div></a>
            </div>
            <!--Excerpt-->
            <a href="" class="pink-text">
              <h6>
                <i class="fa fa-map"></i><strong><?php echo $user['city'];?>, <?php echo $user['state'];?></strong><br>
                <small><?php echo $user['school'];?></small>
              </h6>
            </a>
            <h4><?php echo $user['type'];?></h4>
            <p>
              by <a><strong><?php echo $user['room_name'];?></strong></a>
              <?php echo $user['created'];?>
            </p>
            <label class="badge badge-danger"> non-verified </label>
            <a href="contact.php">
              <label class="badge badge-primary">Report Scam</label>
            </a>
            <p><?php echo $user['discription'];?></p>
            <a class="btn btn-pink btn-rounded">Contact <?php echo $user['room_name'];?></a>
          </div>
        <?php
        } ?>  
      </div>      
    </div>
    <?php
  }
}
else{
  echo(mysql_error());
}

从下拉列表中选择php选项

2 个答案: