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回复讨论(解决方案)
Home Backend Development PHP Tutorial ajax输出的数据,如何让另外一个页面获取到。如何通过表单提交获取到的chk的值。

ajax输出的数据,如何让另外一个页面获取到。如何通过表单提交获取到的chk的值。

Jun 23, 2016 pm 02:24 PM

本帖最后由 qq914260102 于 2013-10-16 10:16:03 编辑

<script type="text/javascript">$(function(){        $("#title").blur(function(){                    $.ajax({                                    url:"check.php",                 type:"GET",                 data:"title="+$("#title").val(),                 success: function(data)                {                      $("#chk").html(data);                 }                                });        })    })</script>  <table>  <form id="reg" name="form1" method="post" action="ser.php" >    <tr>      <td >标题内容:</td>      <td ><input type="text" name="title" id="title"/></td>    </tr>    <tr>      <td >通过输出标题后获取的的:</td>      <td id="chk"></td>    </tr></form>
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上面页面保存为test.php 

请问如何在ser.php 获取到id=“chk”输出的内容。。
需要在test.php页面加入什么内容,在ser.php怎么才能获取到。

回复讨论(解决方案)

将 id=“chk” 的值放入到一个隐藏域中。然后提交php就能获取到。

将 id=“chk” 的值放入到一个隐藏域中。然后提交php就能获取到。


这样子,我试验了一下,获取不到呢。

为input赋值用val() , 且确认一个页面中id必须唯一。

为input赋值用val() , 且确认一个页面中id必须唯一。
领导,可以帮我写一个代码吗,我不会写。。
还有一个问题,就是我标题内容里面输入标题后,通过ajax会自动查询出相关的标题,查询出来的显示通过id=chk显示的。显示出来的内容,是有复选框的,我可以勾选部分,随着标题内容一起提交。。我现在遇到的问题是id=chk显示的内容,我勾选部分后提交,如何获取到这个勾选的值。

<script> <br /> $(function(){ <br /> <br /> $("#title").blur(function(){ <br /> <br /> $.ajax({ <br /> <br /> url:"check.php", <br /> type:"GET", <br /> data:"title="+$("#title").val(), <br /> success: function(data) <br /> { <br /> $("#chk").html(data); <br /> $("#chk_1").val(data); <br /> } <br /> <br /> }); <br /> }) <br /> <br /> }) <br /> </script>
 
  


  
    
    
      
      
    
 
    
      
      
    

那就把data写入到复选框里面去

那就把data写入到复选框里面去
追加了三十分,请领导帮忙。

你的html代码没有贴全

就是刚刚开始的代码。

   


      
      
    

刚刚我的截图,是截取的id="chk"显示的值。我另外一个ajax处理页面check.php输出的结果就是那样子的。
只是我勾选了2和4.。
这个是全部代码了。

我现在就想实现的是,通过ajax查询出相关的内容,然后可以对内容勾选部分和标题一起提交。

现在是把chk得到的结果全部一起提交了。。

每次请求有几个复选框? 复选框的代码你并没有贴出来啊。

对于 $("#chk").html(data); 
你应该贴出 data 的值!
如果 data 中含有表单变量,那么就不会不被提交
使用隐藏域的原因是,你并没有说明是表单元素

对于 $("#chk").html(data); 
你应该贴出 data 的值!
如果 data 中含有表单变量,那么就不会不被提交
使用隐藏域的原因是,你并没有说明是表单元素

?

标题1


标题2


标题3


标题4

每次请求有几个复选框? 复选框的代码你并没有贴出来啊。

每次请求几个复选框看看用户自己的选择,他也可以选择一个,也可以选择多个,也可以不选择,也可以全选。

我在追加一点分,感谢了。

没有被选中的复选框是不会被提交的

没有被选中的复选框是不会被提交的

但是我这样子,怎么才能获取到这几个复选框啊。。。

没有被选中的复选框是不会被提交的

斑竹您没有太明白我的意思,我的这些复选框是从ajax获取过来的,怎么才能把获取过来的部分选择在和其他文本框一起提交到另外页面出来,不是再次提交到那个ajax页面去了。

标题内容:
通过输出标题后获取的的:
通过输出标题后获取的的:
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