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我需要你的幫助來編輯我的腳本,使其輸出結果WHERE title = function(搜索) 如果我輸入:test,它會打印出結果,並將結果作爲標題進行測試。jQuery/MySQL WHERE something = function
search.js:
$('form').submit(function() {
var form_data = ($(this).serialize());
window.location.hash = form_data.replace('=','/');
return false;
});
(function() {
window.App = {
Models: {},
Collections: {},
Views: {},
Router: {}
};
App.Router = Backbone.Router.extend({
routes: {
'': 'index',
'search/:search': 'search',
'*other': 'default'
},
index: function() {
$(document.body).append("");
},
search: function(search) {
$('#result').load('search.php');
}
});
new App.Router();
Backbone.history.start();
})();
的search.php
$query = "SELECT title FROM media WHERE title=";
$result = mysql_query($query);
$row = mysql_fetch_assoc($result);
echo $row['title'];