所以我基于mysql表中的数据生成XML文件,但需要能够使用我的查询从2个表中获取我想要的数据,并且不确定如何执行此操作。详情如下:
TABLE: store_locations
===================================
store_location_id
country
latitude
longitude
===================================
TABLE: store_locations_descriptions
===================================
store_location_id
name
description
city
===================================
PHP功能:
// Function to generate XML file based on store data from database
function fn_store_locator_generate_xml(){
$qur = db_get_field("JOIN QUERY WANTED HERE");
$ans=mysql_query($qur);
$output.= "<?xml version=\"1.0\" encoding=\"UTF-8\"?>
<kml xmlns=\"http://earth.google.com/kml/2.2\">
<Document>";
while($row=mysql_fetch_array($ans))
{
$output.="<name>".$row['name']."</name>";
$output.="<description>".$row['description']."</surname>";
$output.="<Placemark>";
$output.="<name>".$row['name']."</name>";
$output.="<Snippet>".$row['description']."</Snippet>";
$output.="<description>".$row['description']."</description>";
$output.="<Point>";
$output.="<coordinates>".$row['latitude'].",".$row['longitude']."</coordinates>";
$output.="</Point>";
$output.="</Placemark>";
$output.="</person>";
}
$output.="</Document>";
$file_name = "galleries.xml";
$file_pointer = fopen($file_name, "w+");
fwrite($file_pointer, "$output");
fclose($file_pointer);
}
// generate the XML file
fn_store_locator_generate_xml();
非常感谢提前!
答案 0 :(得分:1)
看起来你只需要加入。我假设你在这些表之间有一对一的关系。
select s.store_location_id,
s.country,
s.latitude,
s.longitude,
d.name,
d.description,
d.city
from store_locations s
join store_locations_descriptions d
on s.store_location_id = d.store_location_id
你当然应该研究联接是如何工作的: http://www.keithjbrown.co.uk/vworks/mysql/mysql_p5.php
答案 1 :(得分:0)
SELECT * FROM store_locations JOIN store_locations_descriptions,但你应该看一下mysql join教程,连接对于操纵数据库中的数据至关重要,例如:mysql join tutorial
答案 2 :(得分:0)
SELECT sl.country, sl.latitude, sl.longitude, sld.name, sld.description, sld.city
FROM store_locations sl
JOIN store_location_descriptions sld
ON sl.store_location_id=sld.store_location_id
答案 3 :(得分:0)
试试这个
select s.foo sd.foo
from store_locations s
join store_locations_descriptions sd
on s.store_location_id = sd.store_location_id