使用PHP显示数据库中的随机图像 [英] Display random image from database using PHP

问题描述

我希望显示随机选择的图像或我从数据库中选择的图像。

但我真的不知道如何正确编码。



我尝试了什么： < br $>

 $ query1 = mysqli_query（$ db_con，  select * from image_tb）; 
 while（$ row = mysqli_fetch_array（$ query1））{
  $ img  =  ; 
  $ image_number  = $ row ['  id ]; 
 
 
 
 <   div     class   = 阻止内容崩溃 >  
 <   div     class   =  row-fluid padd-bottom >  
 < span class =code-keyword><   div     class   =  span3 >  
 //我想在这里显示我的数据库上的第一个图像，但我不知道如何编码它prope RLY。 
 <   a    < span class =code-attribute> href   = ＃    class   =  thumbnail >  
 if（1 == $ image_number）{
  <？ php  echo    $ img ; ？>  
} 
 <   / a  >  
 <   / div  >  
 <   div     class   =  span3 >  
 //数据库中的第二张图片
 <   a     href   = ＃    class   =  thumbnail >  
 if（2 == $ image_number）{
 <？ php  echo    $ IMG; ？>  
} 
 <   / a  >  <   / div  >  <   a     href   = ＃    class   = 缩略图 >  
 <   / a  >  <   div     class   =  span3  >  <   a     href   = ＃    class   = 缩略图 >  
 //第3张图片
 <   / a  >  <   a     href   = ＃    class   =  thumbnail > ;  
 if（1 == $ image_number）{
 <？ php  echo    $ img; ？>  
} 
 <   / a  >  
 <   / div  >  
 <   div     class   =  span3 >  
 //第4张图片
 <   a     href   = ＃    class   = 缩略图 >  
 if（1 == $ image_number）{
 < ;  php  echo    $ img ; ？>  
} 
 <   / a  >  
 <   / div  >  
 <   / div  >  
 <？ php 
} 
 ？>  
 <   / div  >  

解决方案

< blockquote> query1 = mysqli_query（

db_con， select * from image_tb< /跨度>）;
while（

row = mysqli_fetch_array（

i want to display image with randomly selected or the one that i choose from database.
but i don't really know how to code it properly.

What I have tried:

$query1=mysqli_query($db_con,"select * from image_tb");
   while($row=mysqli_fetch_array($query1)){
   $img = "";
   $image_number = $row['id'];
   
   

<div class="block-content collapse in">
     <div class="row-fluid padd-bottom">
     <div class="span3">
//i want to display here the first image on my database but i don't how to code it properly. 
      <a href="#" class="thumbnail">
      if (1 == $image_number){
          <?php echo "$img"; ?>
      }
       </a>
       </div>
       <div class="span3">
//the 2nd image from database
      <a href="#" class="thumbnail">
      if (2 == $image_number){
          <?php echo "$img"; ?>
      }
      </a></div><a href="#" class="thumbnail">
       </a><div class="span3"><a href="#" class="thumbnail">
//3nd image 
      </a><a href="#" class="thumbnail">
      if (1 == $image_number){
          <?php echo "$img"; ?>
      }
       </a>
       </div>
<div class="span3">
//4th image 
      <a href="#" class="thumbnail">
      if (1 == $image_number){
          <?php echo "$img"; ?>
      }
       </a>
       </div>
     </div>
<?php
  }
?>
     </div>

解决方案

query1=mysqli_query(

db_con,"select * from image_tb"); while(

row=mysqli_fetch_array(

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