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11行php+js 求改错

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Jun 23, 2016 pm 01:54 PM

如果在document.location后加入百度网址，可以正常实现功能。
或许是js里套php哪里错了？

回复讨论(解决方案)

document.location = "delete1.php?lmbs= &id=";

amp什么意思不太懂，不过像一楼说的去掉以后，依然无法实现

<BODY><input type="button" value="删除" onclick="del()"></BODY><script>function del(){  if (!confirm("您确定要删除这些记录吗？已提交记录将不能删除!")){ return;  }  document.location = "http://baidu.com/"; }</script>

要这样看到你想起了我以前加油
建议你装个火狐 firebug js报错一下就看到了

首先确定页面跳转到delete1.php页面了吗

还有你说的“无法实现”是指什么？

按3楼说法，我把script部分放到了后,现在页面跳转到了delete1.php,但是delete1.php页面没有收到id

$array['id']有值？输出一下。。

amp什么意思不太懂，不过像一楼说的去掉以后，依然无法实现

amp 是 & 这个符号的转义

location 只能跳转不能传值
用 action

function del(){
   if (!confirm("您确定要删除这些记录吗？")){
return;
   }
   document.action = "delete1.php?lmbs=& id=";
}是这样么？
但是，点击删除。弹出对话框，按了“确定”之后就什么反应都没有了。

$array['id']有值？输出一下。。

之前我有用过$select=mysql_query("select * from tb_content",$conn);
$array=mysql_fetch_array($selects)；
刚才加了个echo输出了一下，每个帖子后都有id显示。

什么反应都没有指的是帖子没被删，应该没有转到delete1.php

你在 delete1.php 中 print_r($_GET);
看看都有什么

print_r的结果:失败Array ( [lmbs] => 主题管理 [id] => )

发现lmbs的值传过来了,id没有

我先去考微生物了..我们选修课考试,真的谢谢大家,晚上我再回来研究

那就查看页面“源代码”在 document.location = "delete1.php?lmbs=&id="; 对应的部分都有些什么

我在delete1.php里加了个if($_GET['id'])
显示的结果是没有id被get到

我先去考微生物了..我们选修课考试,真的谢谢大家,晚上我再回来研究

"微生物"....好牛~~ 能不能搞几个作宠物?

function del(){  if (!confirm("您确定要删除这些记录吗？已提交记录将不能删除!")){ return;  }  document.location = "delete1.php?lmbs=<?php echo urlencode("主题管理");?>&id=<?php echo $array['id'];?>";  }

逻辑问题，你的document.location最后得到字符串中的ID没有任何值。。。。

PHP 是服务器端的，JS是客户端的，你的id==$array["id"]?>明显的没有值。这里的id只能通过JS来获取

function del(){  if (!confirm("您确定要删除这些记录吗？已提交记录将不能删除!")){ return;  }else{//这里写点代码用ＪＳ来获取选中的ID,然后用","合并，作为字符串返回// document.location = "delete1.php?lmbs=<?=urlencode("主题管理")?>&id="+id;  } }

alert('delete1.php?lmbs=&id=";');
看对不对。

JS里的值只能通过JS来获取，你这些代码应该是通过ID要删除某条记录，你可以用jquery来写！
<script></script>
<script> <br /> $(document).ready(function(){ <br /> $("#delUserById").click(function del(){ <br /> var id = $(this).parent().find('#userId').val(); <br /> if (!confirm("您确定要删除这些记录吗？已提交记录将不能删除!")){ <br /> return; <br /> } <br /> document.location = "delete1.php?lmbs=<?php echo urlencode("主题管理");?>&id="+id; <br /> }); <br /> }); <br /> </script>

" />

注意导入jquery.js

上面发的还有些问题！
<script></script>
<script> <br /> $(document).ready(function(){ <br /> $("#delUserById").click(function(){ <br /> var id = $(this).parent().find('.userId').val(); <br /> alert(id); <br /> if (!confirm("您确定要删除这些记录吗？已提交记录将不能删除!")){ <br /> return; <br /> } <br /> document.location = "delete1.php?lmbs=<?php echo urlencode("主题管理");?>&id="+id; <br /> }); <br /> }); <br /> </script>

" />

只要你的$array['id']有值，应该可以获取了

[Quote=引用 14 楼 evan2916 的回复:]

我先去考微生物了..我们选修课考试,真的谢谢大家,晚上我再回来研究

"微生物"....好牛~~ 能不能搞几个作宠物?
哈哈，应该伴随我们很久了，而且数量可观。

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