11行php+js 求改错
<script> <br /> function del(){ <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 =<?php echo $array['id']; ?>"; <br /> } <br /> </script>
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
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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
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
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($_GET);
看看都有什么
print_r的结果:失败Array ( [lmbs] => 主题管理 [id] => )
发现lmbs的值传过来了,id没有
我先去考微生物了..我们选修课考试,真的谢谢大家,晚上我再回来研究
那就查看页面“源代码”在 document.location = "delete1.php?lmbs=&id="; 对应的部分都有些什么
那就查看页面“源代码”在 document.location = "delete1.php?lmbs=&id="; 对应的部分都有些什么
我在delete1.php里加了个if($_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?
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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=明显的没有值。这里的id只能通过JS来获取
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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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