我想在两张桌子中搜索。试过this,但没有奏效。我收到错误" Unkown Column ...."
HTML:
if(isset($_POST['button_search']))
{
$search_keyword = $_POST['search_keyword'];
$select = mysql_query("SELECT * from table_1 WHERE match(name_1) against ($search_keyword)
union all SELECT * from table_2 WHERE match(name_2) against ($search_keyword)");
while($row = mysql_fetch_array($select))
{
echo $row['question'];
echo $row['category_name'];
}
}
MYSQL SELECT和PHP:
+-------+-------------+
| id_1 | name_1 |
+-------+-------------+
| 1 | Phinoy |
| 2 | Go |
+-------+-------------+
TABLES:
例如:
表1:
表名:table_1
值(表1):
+-------+-------------+
| id_2 | name_2 |
+-------+-------------+
| 1 | Gi |
| 2 | Phinas |
+-------+-------------+
表2:
表名:table_2
值:
$('.plsCheck').each(function(index, obj){
var fillIn=false;
$(obj).find('select input').each(function(idx, elem){if ($(elem).value !='') fillIn=true;});
if(!fillIn) $(obj).remove();
});
如果我搜索" P"或者" Ph" ,它将显示价值观#Phinoy和Phinas"。
答案 0 :(得分:1)
好。首先,您需要使用INNER JOIN同时比较两个表。在这段代码中,我们正在连接表1和表2的结果,然后我们调用table_1并使用INNER JOIN我们将比较table_2,如果在table_1中有任何结果LIKE table_2并且只发生在列name_1和name_2。
<?php
$query = "SELECT
CONCAT(table_1.name_1,' and ',table_2.name_2)
FROM table_1 INNER JOIN table_2
ON table_1.name_1 LIKE table_2.name_2";
mysql_query($query);
?>