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);
?>
最佳答案
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'";)
关于Mysql 查询在 SQLyog 中工作,但不被浏览器读取,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/39001838/