Question

如果mysql表中的sold = 0，我需要数据具有不同的样式当我使用下面的代码时，网站显示一个空白的白页

（？= $ vehicle-＆gt;这是车辆参考和sold是带有sql表的列

<?php
 if (?=$vehicle->sold?!= 1)                     
    {
    <div class="foo">
       <div class="fboverlay"></div>
       <a>
         <a href="/listing/<?=$vehicle->vehicle_id?>"><img src="/media.php?productId=<?=$vehicle->vehicle_id?>&file=<?=$vehicle->main_image?>" /></a>

       </a>
    </div>
    }        
     else {
          <a href="/listing/<?=$vehicle->vehicle_id?>"><img src="/media.php?productId=<?=$vehicle->vehicle_id?>&file=<?=$vehicle->main_image?>" /></a>       
    }
    ?>

Answer 1

您需要从基础知识中学习PHP。了解运营商，PHP和HTML secion等。无论如何，我修复了你的代码。条件是如果$vehicle->sold不等于1.但我认为，（在您的OP中，您提到它应该为0）您想要这样：$vehicle->sold == 0

//Use sytnax like this. See php opeartors.
if ($vehicle->sold != 1) {
    ?> <!-- Close the php -->
    <div class = "foo">
        <div class = "fboverlay"></div>
            <a href = "/listing/<?= $vehicle->vehicle_id ?>"><img src = "/media.php?productId=<?= $vehicle->vehicle_id ?>&file=<?= $vehicle->main_image ?>" /></a>
    </div>
    <?php
    //Open the php again
} else {
    ?> <!-- close the php -->
    <a href = "/listing/<?= $vehicle->vehicle_id ?>"><img src = "/media.php?productId=<?= $vehicle->vehicle_id ?>&file=<?= $vehicle->main_image ?>" /></a>
<?php //Open again
}

如果来自SQL，则拉出其他

1 个答案: