我看过很多教程,但是他们很困惑,要做我想做的,我只是不知道如何使用这些教程中的现有东西,并使他们按照我想要的方式工作.
我有一个非常简单的表单,包含一个文本框,标签和一个提交按钮.当用户输入表单时,点击提交,我想使用PHP和ajax(使用jquery)将表单的结果插入到MysqL数据库中.
有人可以告诉我这是如何实现的?只是一些非常基本的东西就是我以后让我开始.任何帮助是赞赏.
谢谢
嗨,这里只是一个简单的例子:
HTML:
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"> <html> <head> <title>Quick JQuery Ajax Request</title> <Meta http-equiv="Content-Type" content="text/html; charset=UTF-8"> <!-- include the jquery lib --> <script type="text/javascript" src="jquery.js"></script> <script type="text/javascript"> var ajaxSubmit = function(formEl) { // fetch where we want to submit the form to var url = $(formEl).attr('action'); // fetch the data for the form var data = $(formEl).serializeArray(); // setup the ajax request $.ajax({ url: url,data: data,dataType: 'json',success: function() { if(rsp.success) { alert('form has been posted successfully'); } } }); // return false so the form does not actually // submit to the page return false; } </script> </head> <body> <form method="post" action="process.PHP" onSubmit="return ajaxSubmit(this);"> Value: <input type="text" name="my_value" /> <input type="submit" name="form_submit" value="Go" /> </form> </body> </html>
process.PHP脚本:
<?PHP function post($key) { if (isset($_POST[$key])) return $_POST[$key]; return false; } // setup the database connect $cxn = MysqL_connect('localhost','username_goes_here','password_goes_here'); if (!$cxn) exit; MysqL_select_db('your_database_name',$cxn); // check if we can get hold of the form field if (!post('my_value')) exit; // let make sure we escape the data $val = MysqL_real_escape_string(post('my_value'),$cxn); // lets setup our insert query $sql = sprintf("INSERT INTO %s (column_name_goes_here) VALUES '%s';",'table_name_goes_here',$val ); // lets run our query $result = MysqL_query($sql,$cxn); // setup our response "object" $resp = new stdClass(); $resp->success = false; if($result) { $resp->success = true; } print json_encode($resp); ?>
请注意,这些都没有被测试.我希望它能帮助你.