我想从我的数据库中创建XML。数据保存在不同的表中。这就是为什么我想从不同的表列中选择,如果表中不存在。
如何编写Select以检查表是否包含我需要的列?
$sql = "SELECT products_quantity FROM ".$config['table_name'], $config['table_name2'].products_description;
这是完整的代码
<?php
//database configuration
$config['mysql_host'] = "localhost";
$config['mysql_user'] = "root";
$config['mysql_pass'] = "root";
$config['db_name'] = "sqltoxml";
$config['table_name'] = "products";
$config['table_name2'] = "products_description";
//connect to host
mysql_connect($config['mysql_host'],$config['mysql_user'],$config['mysql_pass']);
//select database
@mysql_select_db($config['db_name']) or die( "Unable to select database");
$xml = "<?xml version=\"1.0\" encoding=\"UTF-8\"?>";
$root_element = $config['table_name']."s";
$xml .= "<$root_element>";
//select all items in table
$sql = "SELECT products_id FROM ".$config['table_name'], products_description FROM ".$config['table_name2'];
$result = mysql_query($sql);
if (!$result) {
die('Invalid query: ' . mysql_error());
}
if(mysql_num_rows($result)>0)
{
while($result_array = mysql_fetch_assoc($result))
{
$xml .= "<".$config['table_name'].">";
//loop through each key,value pair in row
foreach($result_array as $key => $value)
{
//$key holds the table column name
$xml .= "<$key>";
//embed the SQL data in a CDATA element to avoid XML entity issues
$xml .= "<![CDATA[$value]]>";
//and close the element
$xml .= "</$key>";
}
$xml.="</".$config['table_name'].">";
}
}
//close the root element
$xml .= "</$root_element>";
//send the xml header to the browser
header ("Content-Type:text/xml");
//output the XML data
echo $xml;
$fn= "export.xml";
$fp = fopen($fn,"wb");
$write = fwrite($fp,$xml);
fclose($fp);
?>
答案 0 :(得分:0)
这两个表的架构是什么?如果table2在table1中引用了产品ID,则可以连接这两个表。利用IF来指定如果一个字段包含一个值,则返回一个字段,或者oterhwise另一个字段。
SELECT IF(table1.product_quantity,table1.product_quantity,table2.product_description)
FROM table1
JOIN table2 ON table1.id = table2.prodct_id;
这是一种关系数据库的工作方式,允许用户从多个表中获取数据。除非提供其他匹配条件,否则必须存在一些匹配(在列上)以指示您要从另一个表中调用与第一个表中的行相关的行。