PHP：避免将同一产品显示两次？

我正在使用以下代码使用PHP/MYSQL为我的站点创建一个搜索函数

如果选中复选框，此代码将搜索尺寸和颜色，并在“我的页面”中显示它们

例如：

checkbox 1 = small
checkbox 2 = large
checkbox 3 = xxlarge
checkbox 4 = red
checkbox 5 = black
etc etc ......

如果用户选择的

大小为“小”和“大”，颜色为“红”

，my PHP将返回具有这些凭据的产品

这是PHP代码：

<?php
error_reporting(-1);
ini_set('display_errors', 'On'); 
include "config/connect.php";
$searchList = "";
$clause = " WHERE ";//Initial clause
$sql="SELECT *
FROM `yt`
INNER JOIN `ATTRIBUTES` ON yt.id=ATTRIBUTES.id";//Query stub
if(isset($_POST['submit'])){
    if(isset($_POST['keyword'])){
        foreach($_POST['keyword'] as $c){
            if(!empty($c)){
                ##NOPE##$sql .= $clause."`".$c."` LIKE '%{$c}%'";
                $sql .= $clause . " (ATTRIBUTES.sizes LIKE BINARY '$c' OR ATTRIBUTES.colors LIKE BINARY '$c')";
                $clause = " OR ";//Change  to OR after 1st WHERE
            }
        }
        //print "SQL Query: $sql<br />"; //<-- Debug SQl syntax.
        // Run query outside of foreach loop so it only runs one time.
        $query = mysqli_query($db_conx, $sql);
        //var_dump($query); //<-- Debug query results.
        // Check that the query ran fine.
        if (!$query) {
            print "ERROR: " . mysqli_error($db_conx);
        } else {
            // Use $query inside this section to make sure $query exists.
            $productCount = mysqli_num_rows($query);
            $i=0; // count the output amount
            if ($productCount > 0) {
                while($row = mysqli_fetch_array($query, MYSQLI_ASSOC)){
                   $id = $row["id"];
                   $product_name = $row["product_name"];
                   $searchList .= ''.$product_name.'';
                }
            }
        }
    }
}
?> 

但这会引发以下错误：

ERROR: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'INNER JOIN `ATTRIBUTES` ON yt.id=ATTRIBUTES.id WHERE (ATTRIBUTES.sizes LIKE BIN' at line 3

您需要使用SQL“分组依据”函数。如果yt.id是您的产品id，那么将是附加到查询中的“groupbyyt.id”

尝试：

尝试

选择不同的*…

@ᵈˑ，不，那不行。它仍然返回重复的项。因为“项目”表与“颜色和大小”表是分开的。对不起，我应该在我的问题中提到这一点。同样的错误。事实上，在我尝试另一种方法之前，我先像你一样尝试了。两者都抛出相同的错误。

ERROR: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'INNER JOIN `ATTRIBUTES` ON yt.id=ATTRIBUTES.id WHERE (ATTRIBUTES.sizes LIKE BIN' at line 3

<?php

error_reporting(-1);
ini_set('display_errors', 'On');
include "config/connect.php";
$searchList = "";
$clause = " WHERE ";//Initial clause
$sql="SELECT *
FROM `yt`
INNER JOIN `ATTRIBUTES` ON yt.id=ATTRIBUTES.id";//Query stub
if(isset($_POST['submit'])){
    if(isset($_POST['keyword'])){
        foreach($_POST['keyword'] as $c){
            if(!empty($c)){
                ##NOPE##$sql .= $clause."`".$c."` LIKE '%{$c}%'";
                $sql .= $clause . " (ATTRIBUTES.sizes LIKE BINARY '$c' OR ATTRIBUTES.colors LIKE BINARY '$c')";
                $clause = " OR ";//Change  to OR after 1st WHERE
            }
        }

        //append "GROUP BY"
        $sql .= " GROUP BY yt.id ";

        //print "SQL Query: $sql<br />"; //<-- Debug SQl syntax.
        // Run query outside of foreach loop so it only runs one time.
        $query = mysqli_query($db_conx, $sql);
        //var_dump($query); //<-- Debug query results.
        // Check that the query ran fine.
        if (!$query) {
            print "ERROR: " . mysqli_error($db_conx);
        } else {
            // Use $query inside this section to make sure $query exists.
            $productCount = mysqli_num_rows($query);
            $i=0; // count the output amount
            if ($productCount > 0) {
                while($row = mysqli_fetch_array($query, MYSQLI_ASSOC)){
                    $id = $row["id"];
                    $product_name = $row["product_name"];
                    $searchList .= ''.$product_name.'';
                }
            }
        }
    }
}
?>