ajax子页面，提交form表单问题

[clover_lu]

我用ajax做了一个页面gfz.php，其中有个div叫localfresh，点击某按钮用ajax，在localfresh这个div里加载一个页面叫gncreate.php，然后实现局部刷新加载页面。其中gncreate是一个表单页面。加载gncrate.php后此页面的URL还是gfz.php

现在要用gncreate提交表单，简化一下：

<form name="cateform" method="post" id="cateform" action="controller.php">
  <table class="att-tab" >
                <tbody >
                    <tr>
                      <td>高权重：</td>
                      <th><input type="text" class="text-sty" placeholder="请输入高权重" required name="weighth" id="weighth"></th>
                    </tr>
                    <tr>
                      <td>低权重：</td>
                      <th><input type="text" class="text-sty" placeholder="请输入低权重" required name="weightl" id="weightl"></th>
                    </tr>
                    <tr>
                      <td>TAG:</td>
                      <<th><input type="text" class="text-sty" placeholder="请输入TAG" required name="tag" id="tag"></th>
                    </tr>
                     <button type="submit"  name="submit" id="submit" class="btn btn-success" >保存</button>
</form>

其中controller.php里是（节选）

  <?php
ob_start();
session_start();
?>
<?php 
//检测是否登录，若没登录则转向登录界面
if(!isset($_SESSION['username'])){
    header("Location:login.php");
    exit();
}
?>
<?php
include "conn2.php";
if(isset($_POST['submit']))
{
	
	$weighth=$_POST['weighth'];
	$weightl=$_POST['weightl'];
	$tag=$_POST['tag'];
	 $pasql=mysql_query("SELECT * FROM `chapter` where category='$category' and chapter='$chapter' and chapterlist='$chapterlist'");
   $rowpa=Mysql_num_rows($pasql);
		if($rowpa>0){
		echo '<script>alert("此章节已存在！若修改请返回到总攻略页面修改！");window.location="gfz.php";</script>';
		}
		else{
		$reatten=mysql_query("INSERT INTO `attention` (`category`,`chapter`,`chapterlist`,`weighth`,`weightl`,`tag`,`npcs`,`needs`,`npcg`,`needg`,`mainattention`,`skillstate`) VALUES ('$category','$chapter','$chapterlist','$weighth','$weightl','tag','$npcs','$needs','$npcg','$needg','$mainattention','$skillstate');");

		echo '<script>alert("保存成功！");window.location="gfz.php";</script>';
        
		}
   }
   
?>

我现在只能在控制器即controller.php里通过window.location跳转到gfz.php页面，但是无法跳转到含有gncreate.php的gfz.php页面。
该怎样跳到含有gncreate.php的gfz.php页面？或者不刷新直接提交数据到数据表，含有gncreate.php的gfz.php页面不刷新？

[xuzuning]

使用 ajax 就不要使用跳转，否则 ajax 的优势荡然无存

[clover_lu]

引用 1 楼 xuzuning 的回复:

使用 ajax 就不要使用跳转，否则 ajax 的优势荡然无存

那怎样不刷新页面传递数据呢

[yetian26]

不用，你返回的时候用echo 返回，到前台，然后根据返回的值再来判断跳转到哪个页面，而不是在控制器里跳转

[yetian26]

然后if(isset($_POST['submit']))
{

$weighth=$_POST['weighth'];
$weightl=$_POST['weightl'];
$tag=$_POST['tag'];
$pasql=mysql_query("SELECT * FROM `chapter` where category='$category' and chapter='$chapter' and chapterlist='$chapterlist'");
$rowpa=Mysql_num_rows($pasql);
if($rowpa>0){
echo 'yes';
}
else{
$reatten=mysql_query("INSERT INTO `attention` (`category`,`chapter`,`chapterlist`,`weighth`,`weightl`,`tag`,`npcs`,`needs`,`npcg`,`needg`,`mainattention`,`skillstate`) VALUES ('$category','$chapter','$chapterlist','$weighth','$weightl','tag','$npcs','$needs','$npcg','$needg','$mainattention','$skillstate');");

echo 'no';

}
}

?>

[xiaozhanger]

引用 4 楼 yetian26 的回复:

然后if(isset($_POST['submit'])) { $weighth=$_POST['weighth']; $weightl=$_POST['weightl']; $tag=$_POST['tag']; $pasql=mysql_query("SELECT * FROM `chapter` where category='$category' and chapter='$chapter' and chapterlist='$chapterlist'"); $rowpa=Mysql_num_rows($pasql); if($rowpa>0){ echo 'yes'; } else{ $reatten=mysql_query("INSERT INTO `attention` (`category`,`chapter`,`chapterlist`,`weighth`,`weightl`,`tag`,`npcs`,`needs`,`npcg`,`needg`,`mainattention`,`skillstate`) VALUES ('$category','$chapter','$chapterlist','$weighth','$weightl','tag','$npcs','$needs','$npcg','$needg','$mainattention','$skillstate');"); echo 'no'; } } ?>

这个是正解

[此人怕不是个傻子呦]