我尝试发送搜索查询并通过ajax显示结果。出于某种原因,当我提交搜索时,我在控制台中收到404 Not Found错误消息。任何提示或指示将不胜感激!
更新:只是为了澄清网址不是foo / local / bar / ...,我只是把这些东西保密,因为我是偏执的大声笑。路线被要求,这是唯一涉及的路线:
/*Home Page*/
Route::get('/', array(
'as' => 'home',
'uses' => 'HomeController@home'
));
以下是控制台错误日志:
POST http://foo/local/bar/public/ 404 (Not Found)
jquery.js:9597
Object
{readyState: 4, getResponseHeader: function, getAllResponseHeaders: function,
setRequestHeader: function, overrideMimeType: function…} home.js:137
error home.js:138
Not Found home.js:139
JS:
$(document).ready(function(){
$("#search_form").submit(function(e) {
e.preventDefault();
//form_data
var data = $('#search_form').serializeArray();
$.ajax({
url: window.location,
type: "POST",
data: data,
success: function(data){
$("#search_results").html($search);
},
error: function(xhr, status, error){
console.log(xhr);
console.log(status);
console.log(error);
$("search_results").html("Failure");
}
});
});
});
控制器:
public function post_index() {
$search_table = Input::get('search_table');
$search_column = Input::get('search_column');
$search_input = Input::get('search_input');
$search = DB::table($search_table)->where($search_column, 'LIKE', "%{$search_input}%")->get();
echo json_encode($search);
exit;
}
查看:
<form id="search_form" method="POST" action="" >
<select id="search_table" class="search_table">
<option value = ''>Search</option>
<option value = 'commissions'>Search commissions</option>
</select>
<div class="search_box" id="search_column_div">
<select id='search_column' name='search_column'>
<option value='select'>by ...</option>
<option value='street_address'>by street address</option>
</select>
</div>
<div id="search_input_div">
<input id='search_input' name='search_input' class='search_input' placeholder='Enter Street Address'/>
</div>
<input type="submit" class="button expand radius" id="search_button"/>
</form>
答案 0 :(得分:1)
我可以立即看到您的代码存在两个问题。您的路由正在侦听GET请求,但您在AJAX调用中使用POST请求。第二个问题是您已经错误地命名了控制器功能。您正在使用较旧的Laravel 3语法,但您已在问题中标记了Laravel 4。使用Laravel 4 RESTful控制器函数命名约定,您的函数名称应为postIndex()