Question

我想在没有图像时隐藏我的html div，并在我有图像时显示它。我怎么能用PHP做到这一点？这是我的代码

<?php
   $imgsql=mysqli_query($link,"select eve_pic,eve_pic2,eve_pic3,eve_pic4 from event_tbl");

?>
                <div class="col-md-3 col-sm-6 col-xs-12" <?php  if(mysqli_num_rows($imgsql)>0){ 
                      $imgrow=mysqli_fetch_array($imgsql);
               ?> style="visibility:visible;"<?php }?> <?php else { ?> style="display:none" <?php}?> >
                    <div class=sz_portfolio_003_trig_wrap>
                              <div class=sz_portfolio_003_trig_img>
                        <div class=sz_portfolio_003_trig_btn>
                                 <a href=#sz_portfolio_003_4_col data-backdrop=true data-slide-to=0 data-toggle=modal>
                            <img src="admin/uploads/eventpics/<?php echo $imgrow['eve_pic'];?>" alt="sz_portfolio_003_01" width="200px" height="200px" class="imagen"/></a>

                            </div>
                        </div>

                    </div>
                </div>

Answer 1

您需要使用If语句来获得所需内容。要了解有关If语句的更多信息，请阅读here

改变这个：

UIEdgeInsets

对此：

    <?php
   $imgsql=mysqli_query($link,"select eve_pic,eve_pic2,eve_pic3,eve_pic4 from event_tbl");

?>
                <div class="col-md-3 col-sm-6 col-xs-12" <?php  if(mysqli_num_rows($imgsql)>0){ 
                      $imgrow=mysqli_fetch_array($imgsql);
               ?> style="visibility:visible;"<?php }?> <?php else { ?> style="display:none" <?php}?> >
                    <div class=sz_portfolio_003_trig_wrap>
                              <div class=sz_portfolio_003_trig_img>
                        <div class=sz_portfolio_003_trig_btn>
                                 <a href=#sz_portfolio_003_4_col data-backdrop=true data-slide-to=0 data-toggle=modal>
                            <img src="admin/uploads/eventpics/<?php echo $imgrow['eve_pic'];?>" alt="sz_portfolio_003_01" width="200px" height="200px" class="imagen"/></a>

                            </div>
                        </div>

                    </div>
                </div>

Answer 2

            <div class="col-md-3 col-sm-6 col-xs-12" style="<?php 

                  if (mysqli_num_rows($imgsql)>0){ 
                  $imgrow=mysqli_fetch_array($imgsql);
                  echo "display:block;"}

                  else { echo "display:none"};

             ?>"...

Answer 3

只需将其更改为：

<?php if(!empty($imgrow['eve_pic']))?>{
<div class=sz_portfolio_003_trig_wrap>
    <div class=sz_portfolio_003_trig_img>
        <div class=sz_portfolio_003_trig_btn>
            <a href=#sz_portfolio_003_4_col data-backdrop=true data-slide-to=0 data-toggle=modal>
            <img src="admin/uploads/eventpics/<?php echo $imgrow['eve_pic'];?>" alt="sz_portfolio_003_01" width="200px" height="200px" class="imagen"/></a>

        </div>
    </div>
</div>
<?php } ?>

如果图像存在，则只显示div。

我想使用php

3 个答案: