我试图创建一个PHP MySQL的搜索引擎的工作就像用ajax自动建议一封comerce搜索引擎吗?......我的表像如何将自动建议添加到我的php mysql搜索引擎?
id cat name
1 men subi
2 men flick
3 women sheeba
4 women leena
我的形式是像
<html>
<head>
<title>search engine</title>
</head>
<body>
<form action = 'ss.php' method ='GET'>
<input type = "text" name = "q">
<input type = "submit" name = "submit" value = "search"
</body>
</html>
我ss.php是
$k = $_GET["q"];
$con = mysqli_connect("localhost", "root", "");
mysqli_select_db($con,"x");
$terms=explode(" ",$k);
$i=0;
$set_limit = ("9");
$subi = "";
foreach ($terms as $each)
{
$i++;
$escapedSearchString = mysqli_real_escape_string($con,$each);
if ($i == 1)
$subi.= " title LIKE '%$escapedSearchString%' ";
else
$subi.= " AND title LIKE '%$escapedSearchString%' ";
}
$query = "select SQL_CALC_FOUND_ROWS * from table WHERE $subi order by rand() limit $set_limit";
$qry = mysqli_query($con,"$query");
$row_object = mysqli_query($con,"Select Found_Rows() as rowcount");
$row_object = mysqli_fetch_object($row_object);
$actual_row_count = $row_object->rowcount;
$result = $actual_row_count;
它的做工精细当我搜索一个词像SUBI或sheeba BT我要的是,如果我开始键入一个字的“它“会告诉像
sheeba
subi
sheeba in women
subi in men
中国汽车的建议,如果用户点击sheeba查询机器会自动变为喜欢这个
" select * from table where title like '%sheeba%' "
,如果用户点击了“女性sheeba”的“LL查询改为喜欢这个
" select * from table where cat = 'women' and title like '%sheeba%' "
我怎样才能得到这个? 请简要回答... TNX提前....
为此使用'ajax'并显示响应的建议。 –