<?php
$connect = mysqli_connect("localhost","root","","test");

if (mysqli_connect_errno())
        {
            echo "Failed to connect to MySQL: " . mysqli_connect_error();
        }
$q = mysqli_query($connect,"SELECT * FROM feed");                                                    
$fetch1 = mysqli_fetch_all($q);
$fetch2=$fetch1;

$fn1= count($fetch1);
$fn2=count($fetch2);

for($i=0;$i<$fn1;$i++)
    {        
        for($j=0;$j<$fn2;$j++)

        {
            if( $fetch1[$i][3] == $fetch2[$j][3])
                {
                    echo "duplicat";
                   echo "<div class='post'>";
                         print($fetch1[$i][1]);   print($fetch1[$i][2]); 
                    echo  "</div><br>";  

                    break 2;
                }
        }
    }

?>

尝试此查询这将按代码分组文件

$q = mysqli_query($connect,"SELECT GROUP_CONCAT(file, ' ') AS files,* FROM feed GROUP BY code");
 $fetch = mysqli_fetch_array($q);
 while($fetch = mysqli_fetch_array($q)) {

 echo '<div class="post">'.$fetch['feed'].' '.$fetch['files'].'</div><br>';

 }
 ?>

有关group\u concat的更多信息，请参见此处group\u concat逗号分隔符-mysql

最简单的解决方案是将查询更改为使用 GROUP BY 做一个 GROUP CONCAT 在 FILE :

$q = mysqli_query($connect,"SELECT FEED, GROUP_CONCAT(FILE) AS file, CODE FROM feed GROUP BY CODE");

这将给你一行每 CODE 输出值，如下所示：

FEED        FILE                                                         CODE
Test post   uploads/1.jpg, uploads/2.jpg, uploads/3.jpg, uploads/4.jpg   54231

如果你想要一个除 , 例如。 / ，更改 GROUP_CONCAT 到

GROUP_CONCAT(FILE SEPARATOR '/')

rextester上的演示
更新
php代码中还有一个错误，您正在设置 $fetch = mysqli_fetch_array($q) 在while循环之前，然后不使用它做任何事情，因此会丢失一行值。请尝试以下代码：

$q = mysqli_query( $connect, "SELECT FEED AS feed, GROUP_CONCAT(FILE) AS file, CODE FROM feed GROUP BY CODE" );
 while ( $fetch = mysqli_fetch_array( $q ) ) {
    echo '<div class="post">' . $fetch[ 'feed' ] . $fetch[ 'file' ] . '</div><br>';
 }