我有一份搜索表:
<form action="search.php" method="post" name="form" id="form" onsubmit="return false;">
<input name="search" type="text" id="search">
<input name="name" type="text" id="name">
<select name="car" id="car">
<option value="audi">AUDI</option>
<option value=bmw">BMW</option>
<option value="mini">MINI</option>
</select>
<select name="year" id="year">
<option value="2000">2000</option>
<option value="2005">2005</option>
<option value="2006">2006</option>
</select>
</form>
和search.php
<?php
$search = $_POST['search'];
$search = addslashes($search);
$search = htmlspecialchars($search);
$search = stripslashes($search);
$year = $_POST['option'];
$year = addslashes($year);
$year = htmlspecialchars($year);
$year = stripslashes($year);
if($search == '')
exit("Enter name");
elseif(!preg_match("/^[a-zа-я0-9]+$/ui", $search))
exit("Incorrectly");
include 'dbsetting.php';
$mysqli->set_charset("utf8");
$query = "SELECT * FROM cars WHERE (type LIKE '%".$search."%' OR name LIKE '%".$search."%' OR class LIKE '%".$search."%') AND year = '".$year."'";
现场搜索成功了。使用AJAX代码:
$(function() {
$("#search").keyup(function(){
var search = $("#search").val();
$.ajax({
type: "POST",
url: "search.php",
data: {"search": search},
cache: false,
success: function(response){
$("#res").html(response);
}
});
return false;
});
});
输入name=“搜索”正确工作,但我不能添加到搜索另一个选择和更多的输入。如何在ajax代码中添加一些输入和选择,以及如何在search.php中保存和添加搜索结果?
我为我的英语感到抱歉。
发布于 2014-02-24 05:25:01
您可以张贴您的字段的名称值对格式,如:-
var search = $("#search").val();
var name = $("#name").val();
var car = $("#car").val();
var year = $("#year").val();
以及用户发送ajax请求作为数据字段时的查询字符串。
data: {"search": search, "name":name, "car": car, "year":year},
也可以使用$(“表单”).serialize();自动将所有窗体字段转换为名称值对(类似)。
var my_data = $( "form" ).serialize();
// user that variable in your ajax call.
data: my_data,
发布于 2014-02-24 04:41:54
尝试一下,使用jquery
$('#search, #car, #year').bind( "keyup change", function() {
//put your ajax here
});
发布于 2014-02-24 04:45:00
$(function() {
$("#search").keyup(function(){
var search = $("#search").val();
var name = $("#name").val();
var car = $("#car").val();
var year = $("#year").val();
$.ajax({
type: "POST",
url: "search.php",
data: {"search": search, "name":name, "car": car, "year":year},
cache: false,
success: function(response){
$("#res").html(response);
}
});
return false;
});
});
https://stackoverflow.com/questions/21978959
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