Question

表：tblBuilding
buildingid .......... buildingname ................ buildinglocation
1 ......... ................诊所....................... ........爱丽丝路
2 ......... ................冬季购物中心..................... Damansara
3 ......... ................教师....................... ....北京路 4 ......... ................ Star Mall ...................... ...... Pavillion

表：tblperson
personid ....... personname .................. personno .......... personbuilding < BR /> 1 .................... Andrew Clarkson博士...... 0312883232 .......诊所
2 .................... Antoine先生...................... 0825112315 .... ...冬季商城
3 .................... Emmy Dahliana夫人....... 0112357125 .......教师们 4 ....................贾斯汀安德鲁教授......... 0524128866 .......教职员工 5 .................... Sanila博士....................... 0225113070 ...... ......诊所

如果我输入关键字 Andrew ，我的预期结果为
ID：1个
人名：Andrew Clarkson博士 personno：0312883232
人员：诊所
建筑物位置：爱丽丝路
如果我输入关键字 Star Mall ，我的预期结果为
buildingid：4
buildingname：Star Mall
建筑物位置：Pavillion

我的代码：

if (isset($_GET["FirstName"])) 
{
    $FirstName = $_GET['FirstName'];
}

$query= "(SELECT * FROM tblBuilding where buildingname like '%".$FirstName."%' OR buildinglocation '%".$FirstName."%') UNION (SELECT * FROM tblperson where personname like '%".$FirstName."%' OR personno '%".$FirstName."%' or personbuilding (SELECT * FROM tblBuilding where buildingname like '%".$FirstName."%' OR buildinglocation '%".$FirstName."%')) "

mysql_query($query);

此查询返回空结果。错误如下所示。

   Warning: mysql_num_rows() expects parameter 1 to be resource,
   boolean given in C:\xampp\htdocs\dbinfo\getDetails.php on line 68 {"success":0,"message":"Keyword(s) not found"}

Answer 1

UNION运算符使用相同的列名。您应该为具有相同名称的两个表的列名别名。它应该工作。

Answer 2

我认为现在已经修复了。尝试一下:)注意：请注意，UNION中的每个SELECT语句必须具有相同的列数。列还必须具有类似的数据类型。此外，每个SELECT语句中的列必须具有相同的顺序。

if (isset($_GET["FirstName"])) 
{
$FirstName = $_GET['FirstName'];
}

$query= "(SELECT * FROM tblBuilding where buildingname like '%".$FirstName."%' OR buildinglocation LIKE '%".$FirstName."%' ) UNION (SELECT * FROM tblperson where personname LIKE '%". $FirstName ."%' OR personno '%".$FirstName."%' or personbuilding (SELECT * FROM tblBuilding where buildingname LIKE '%".$FirstName."%' OR buildinglocation '%".$FirstName."%')) ";

 mysql_query($query);

PHP搜索多个表

2 个答案: