该守则给出了错误＆＃34;警告：mysqli_fetch_assoc（）期望参数1为mysqli_result，boolean＆＃34;

Question

代码给出错误，即使我已经多次检查它实际上在数据库中有4个表，并且通过一些操作我将我的最终值放到名为final_list的表中。 如果有人可以帮助我，请

$query_show="SELECT * FROM india";
$result_show = mysqli_query($conn,$query_show);
$pos1=-1;
$pos2=-1;
$pos3=-1;
$pos4=-1;
while($tab1 = mysqli_fetch_assoc($result_show))
{   
    $pos1 = $tab1['id1'];
    $str=$tab1['col1'];
    $query_show2="SELECT * FROM edu WHERE col2 = '{$str}'";
    $result_show2 = mysqli_query($conn,$query_show2);
    while($tab2 = mysqli_fetch_assoc($result_show2))
        {
            $pos2=$tab2['id2'];
            $query_show2 = "DELETE FROM edu WHERE col2 = '{$str}' LIMIT 1";
            $result_show2 = mysqli_query($conn, $query_show2);
        }
    $query_show3="SELECT * FROM eng WHERE col3='{$str}'";
    $result_show3 = mysqli_query($conn,$query_show3);
    while($tab3 = mysqli_fetch_assoc($result_show3))
        {
            $pos3=$tab3['id3'];
            $query_show3 = "DELETE FROM eng WHERE col3 = '{$str}' LIMIT 1";
            $result_show3 = mysqli_query($conn, $query_show3);
        }
    $query_show4="SELECT * FROM out WHERE col4='{$str}'";
    $result_show4=mysqli_query($conn,$query_show4);
    while($tab4=mysqli_fetch_assoc($result_show4))
        {
            $pos4=$tab4['id4'];
            $query_show4 = "DELETE FROM out WHERE col4 = '{$str}' LIMIT 1";
            $result_show4 = mysqli_query($conn, $query_show4);
        }
    $rnks=$pos1." ".$pos2." ".$pos3." ".$pos4;
    $query_insert_final= "INSERT INTO final_list(college,ranks) VALUES ('{$str}','{$rnks}')";
    $result = mysqli_query($conn,$query_insert_final);
    //$query_show = "DELETE FROM india WHERE col2 = '{$str}' LIMIT 1";
    //$result_show = mysqli_query($conn, $query_show);
}

＆GT？; 在