我正在搜索一个我正在为朋友工作的房产网站。在数据库中,属性类型由id编号命名,例如: house = 30,flat = 8,terraced =1,依此类推。
我如何从数据库中检索所有的属性,当一些独立的房子的值是2,房子的值是30等等:)它得到了我的stuck..lol,这是我到目前为止没有工作的…
$bedrooms = $_GET['bedrooms'];
$pricefrom = $_GET['pricefrom'];
$priceto = $_GET['priceto'];
$proptype = $_GET['proptype'];
if($proptype == 'house'){
$search_propsubid = array('1,2,3,4,5,6,21,22,23,24,26,27,30');
}elseif($proptype == 'flat'){
$search_propsubid = array('7,8,9,10,11,28,29,44');
}elseif($proptype == 'bungalow'){
$search_propsubid = array('');
}
$sql = mysql_query("SELECT * FROM `properties` WHERE `PROP_SUB_ID`='$search_propsubid' AND `BEDROOMS`='$bedrooms' AND `TRANS_TYPE_ID`='1' HAVING `PRICE` BETWEEN '$pricefrom' AND '$priceto' ORDER BY `UPDATE_DATE` DESC");谢谢你抽出时间来,我希望有人能给我指路。向Steve致敬
发布于 2012-05-27 09:31:17
你可以尝试implode数组:
$search_propsubid = array('1,2,3,4,5,6,21,22,23,24,26,27,30');
$comma_separated = implode(",", $search_propsubid);
$sql = mysql_query("SELECT * FROM `properties` WHERE `PROP_SUB_ID` in ($comma_separated) ...如果这对你不起作用,请告诉我消息。
发布于 2012-05-27 09:33:56
您可以使用MySql IN()比较运算符选择所有与值列表匹配的值:
$sql = mysql_query("
SELECT *
FROM `properties`
WHERE `PROP_SUB_ID` IN (" .implode(",", $search_propsubid). ")
AND `BEDROOMS`='$bedrooms'
AND `TRANS_TYPE_ID`='1'
HAVING `PRICE` BETWEEN '$pricefrom' AND '$priceto'
ORDER BY `UPDATE_DATE` DESC
");假设使用$proptype == 'flat',则输出将为:
SELECT *
FROM `properties`
WHERE `PROP_SUB_ID` IN (7,8,9,10,11,28,29,44)
...https://stackoverflow.com/questions/10771008
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