php - jquery不会用post方法发送

Question

$.post 没有发送任何内容

     function clicked()
 {
   var $contact_title=$("#contact_title");
   var $contact_summary=$("#bbcode");
   alert($contact_title.val());// How do I get the contents of the title
   alert($contact_summary.val());// How do I get the contents of the textarea
   $.post('jquery_send_admin.php',{ title:$contact_title, content:$contact_summary }, function(data){ alert("Message was sent") }, 'html');
 }

我在控制台错误中遇到异常..如下所示:

更新:下一页没有插入数据..为什么？!？

     if( isset($_POST["title"]) && isset($_POST["content"]) )
  {
      $title=mysql_escape_string($_POST["title"]);
      $content=mysql_escape_string($_POST["content"]);

       $result=mysql_query("INSERT INTO users (query_title,query_message) VALUES(''$title', '$content')") or die(mysql_error());

   }

发生以下错误:

Error: uncaught exception: [Exception... "Could not convert JavaScript argument"  nsresult: "0x80570009 (NS_ERROR_XPC_BAD_CONVERT_JS)"  location: "JS frame :: http://ajax.googleapis.com/ajax/libs/jquery/1.5/jquery.min.js :: <TOP_LEVEL> :: line 16"  data: no]

更新:这就是我从页面请求的内容，它是由 jquery 触发的:

    <?php
echo 'outside';
  if( isset($_POST["title"]) && isset($_POST["content"]) )
  {
      echo 'inside';
      $title=mysql_escape_string($_POST["title"]);
      $content=mysql_escape_string($_POST["content"]);

       $result=mysql_query("INSERT INTO users (query_title,query_message) VALUES(''$title', '$content')") or die(mysql_error());

   }

?>

Answer 1

您需要使用 .val() 提取值方法:

var $contact_title = $('#contact_title').val();
var $contact_summary = $('#bbcode').val();
var dataToPost = { title: $contact_title, content: $contact_summary };
$.post('jquery_send_admin.php', dataToPost, function(data) { 
    alert('Message was sent'); 
}, 'html');