Mysql 查询在 SQLyog 中工作，但不被浏览器读取

Question

这段代码在mysql查询中有什么错误？？？我认为这是一个非常简单的问题，但找不到错误:(虽然这个 mysql 查询在 SQLYog 中工作

$result =mysql_query(SELECT NUMPOINTS((i.geom))COUNT FROM district i WHERE district_name="D.G.KHAN";)

比赛代码是

//$type="R";
$result =mysql_query(SELECT NUMPOINTS((i.geom))COUNT FROM district i WHERE district_name="D.G.KHAN";)
$row2 = mysql_fetch_array($result);
$count=$row2['count'];
$x_points = array();
$y_points = array();
/* $row1 ['color']="00000"; */
for($c=1;$c<=$count;$c++){
    $res=mysql_query("SELECT district_name,X ( POINTN ( i.geom ,$c))xx ,Y( POINTN ( i.geom ,$c))yy ,color FROM district i where dis_id='".$id."'" );
    while($row1 = mysql_fetch_array($res)){
    array_push($x_points['xx']);
    array_push($y_points['yy']);    
 }
}

print_r ($x_points);
print_r ($y_points);


?>

Answer 1

您需要添加为字符串:

$result =mysql_query("SELECT NUMPOINTS((i.geom)) FROM district i WHERE district_name='D.G.KHAN'";)