如果php中的属性等于1，则隐藏div

Question

用户单击查看详细信息后，我将显示特定属性的属性详细信息。如果属性类型值为 Open Plots，那么我不想显示停车场 div。我为此编写了代码，但出现意外错误 '< '这是我写的代码：<?php session_start();include 'db.php';   $id = (int)$_GET['id'];$sql = "SELECT * FROM tbl_properties WHERE property_id = $id";$oppointArr =array();$result = mysqli_query($conn,$sql);if (mysqli_num_rows($result) > 0) {  while($row = mysqli_fetch_array($result))   {              $oppointArr = $row;             }}else {  echo "0 results";}?>        <input  type='hidden' value='<?=$id;?>' name='property_id'>    <div class="property-specs">                            <ul class="specs-list">                                <li><div class="icon"><span class="flaticon-double-king-size-bed"></span></div> <?php echo $oppointArr['property_type'];?></li>                                <li><div class="icon"><span class="flaticon-copy"></span></div> <?php echo $oppointArr['area_sqft'];?> Sqft</li>                                <li><div class="icon"><span class="fa fa-compass"></span></div> <?php echo $oppointArr['facing'];?> Facing</li>                                <li id="car_parking"><div class="icon"><span class="flaticon-private-garage"></span></div>                                <?php if((<?php echo $oppointArr['property_type'];?>)=='Open-Plots')                                {                                    <?php echo 'style="display:none;"';?>                                }                                else{                                <?php echo $oppointArr['car_parking'];?>                                }                                </li>                                <li id="total_bathrooms"><div class="icon"><span class="flaticon-vintage-bathtub"></span></div> <?php echo $oppointArr['total_bathrooms'];?> Bathrooms</li>                            </ul>                        </div>

Answer 1

使用以下内容更改您的代码条件。

<?php session_start();

include 'db.php';   

$id = (int)$_GET['id'];

$sql = "SELECT * FROM tbl_properties WHERE property_id = $id";

$oppointArr =array();

$result = mysqli_query($conn,$sql);

if (mysqli_num_rows($result) > 0) 

{

  while($row = mysqli_fetch_array($result)) 

  {          

    $oppointArr = $row;           

  }

}

else 

{

  echo "0 results";

}

$display = 'block';

if(($oppointArr['property_type'])=='Open-Plots'){

    $display = 'none';

}

?>

<input  type='hidden' value='<?=$id;?>' name='property_id'>

<div class="property-specs">

    <ul class="specs-list">

        <li><div class="icon"><span class="flaticon-double-king-size-bed"></span></div> <?php echo $oppointArr['property_type'];?></li>

        <li><div class="icon"><span class="flaticon-copy"></span></div> <?php echo $oppointArr['area_sqft'];?> Sqft</li>

        <li><div class="icon"><span class="fa fa-compass"></span></div> <?php echo $oppointArr['facing'];?> Facing</li>

        <li id="car_parking" style="display:<?php echo $display?>">><div class="icon"><span class="flaticon-private-garage"><?php echo $oppointArr['car_parking'];?></span></div></li>

        <li id="total_bathrooms"><div class="icon"><span class="flaticon-vintage-bathtub"></span></div> <?php echo $oppointArr['total_bathrooms'];?> Bathrooms</li>

    </ul>

</div>

Answer 2

<?php 

  if($oppointArr['property_type']=='Open-Plots'){

     echo 'style="display:none;"';

  }else{

     echo $oppointArr['car_parking'];

  }

?>

这种方式更好吗？