Php 比较MYSQL中的两个表

我必须比较两个表，然后将它们列为复选框，而匹配的值应该被选中，而不匹配的值应该被选中

//表1

//表2

从两个表中匹配rid，其中category='uniform' 列出所有并检查匹配的rid

$query = "select a.* from table1 a, table b where a.rid = b.rid and a.category = b.category";

$result = mysqli->query($query);

$fetched = $result->fetch_assoc();

foreach($fetched as $row){
echo "<input type='checkbox' name='{$row['rid']}' value='' checked > {$row['category']}";
}

---

循环浏览表1中的每一行。如果rid存在于表2中，请勾选。我也必须完成我的工作。有志愿者吗？基于rid和类别过滤器在两个表之间进行外部联接。类别值统一标记为勾号，类别为空则不标记。

id|rid | category  
1 | 1  |  uniform  
2 | 2  |  uniform

$query = "select a.* from table1 a, table b where a.rid = b.rid and a.category = b.category";

$result = mysqli->query($query);

$fetched = $result->fetch_assoc();

foreach($fetched as $row){
echo "<input type='checkbox' name='{$row['rid']}' value='' checked > {$row['category']}";
}

    $query = "select t1.rid, 'matched' as matching from table1 t1 where t1.rid  in (
              select rid from table2 where category = 'uniform')
             union
             select t1.rid ,'notmatched' from table1 t1 where t1.rid not in (
             select rid from table2 where category = 'uniform')";
    $result = mysqli->query($query);
    $fetched = $result->fetch_assoc();
    foreach($fetched as $row){
    if($row['matching'] = 'matched'){
           echo "<input type='checkbox' name='{$row['rid']}' value='' checked='checked' >"; }
    else{ 
            echo "<input type='checkbox' name='{$row['rid']}' value=''  >"; }
     }