检查发现post为空 应该怎么写
新手一个 刚自学了两天 过程中卡在不少地方过 这里是好不容易各种查各种找才搞出来的代码 现在卡在这里 求助啊
<?php $myconn=mysql_connect("localhost","root",""); mysql_select_db("login_test",$myconn); $strSql="select * from table_test"; $result=mysql_query($strSql,$myconn);?><html> <head> <title>Print test</title> <link href="css.css" rel="stylesheet" type="text/css" /> </head> <body> <br> 显示： <br> <form METHOD="POST" action="<?php echo $_SERVER['PHP_SELF'];?>"> <table id="table1"> <tr class="table_head"> <td>objname</td> <td>objinfo</td> </tr> <?php while($row=mysql_fetch_array($result)) { ?> <tr> <td><?php echo $row['objname']?></td> <td><?php echo $row['objinfo']?></td> <td><input type="hidden" name="id" value="<?php echo $row['id']?>"><input type="submit" value="delete"/></td> </tr> <?php } ?> </table> <div><a href="http://localhost/test2/submit.php">点此添加信息</a></div> </form> <?php echo empty($_POST); /*if(!empty($_POST)){ $id = $_POST['id']; $sql = "delete from student where objid='$id'"; $query = mysql_query($sql); if($query){ echo "<script>alert('删除成功');history.back();</script>"; } }*/ ?> </body><?php mysql_close($myconn); ?></html>
回复讨论(解决方案)
没看到你有submit。
你可以加上print_r($_POST); 看看有没有post
没看到你有submit。
你可以加上print_r($_POST); 看看有没有post
我在29行后面有个submit 用print_r($_POST)查看后是空的Array
提交后，$_POST 不会为空
但 $_POST['id'] 只用一个，并且是最后一个 $row['id']
这显然是不对的
提交后，$_POST 不会为空
但 $_POST['id'] 只用一个，并且是最后一个 $row['id']
这显然是不对的
版主大大
那如果我要给每一行附一个ID 用于之后的定位和删除应该怎么写呢
如果你想通过表单提交来实现删除，那么每行（tr）都应有一个表单（form）
而包围在外面的表单是不需要的
action=""
是提交到本页，可以不写
这样提交后，$_POST['id'] 才是待删除的 id
如果你想通过表单提交来实现删除，那么每行（tr）都应有一个表单（form）
而包围在外面的表单是不需要的
action=""
是提交到本页，可以不写
这样提交后，$_POST['id'] 才是待删除的 id
<?php $myconn=mysql_connect("localhost","root",""); mysql_select_db("login_test",$myconn); $strSql="select * from table_test"; $result=mysql_query($strSql,$myconn);?><html> <head> <title>Print test</title> <link href="css.css" rel="stylesheet" type="text/css" /> </head> <body> <br> 显示： <br> <table id="table1"> <tr class="table_head"> <td>objname</td> <td>objinfo</td> </tr> <?php while($row=mysql_fetch_array($result)) { ?> <form METHOD="POST"> <tr> <td><?php echo $row['objname']?></td> <td><?php echo $row['objinfo']?></td> <td><input type="hidden" name="id" value="<?php echo $row['id']?>"> <input type="submit" value="delete"/></td> </tr> </form> <?php } ?> </table> <div><a href="http://localhost/test2/submit.php">点此添加信息</a></div> <?php echo empty($_POST); print_r($_POST); /*if(!empty($_POST)){ $id = $_POST['id']; $sql = "delete from student where objid='$id'"; $query = mysql_query($sql); if($query){ echo "<script>alert('删除成功');history.back();</script>"; } }*/ ?> </body><?php mysql_close($myconn); ?></html>我改成这样以后 echo的post依然是空啊 是哪出问题了？
你点击 delete 按钮了吗？
贴出浏览器中的 html
<html> <head> <title>Print test</title> <link href="css.css" rel="stylesheet" type="text/css" /> </head> <body> <br> 显示： <br> <table id="table1"> <tr class="table_head"> <td>objname</td> <td>objinfo</td> </tr> <form METHOD="POST"> <tr> <td>obj1</td> <td>this is obj1</td> <td><input type="hidden" name="id" value="<br /><b>Notice</b>: Undefined index: id in <b>C:\xampp\htdocs\test2\test.php</b> on line <b>30</b><br />"> <input type="submit" value="delete"/></td> </tr> </form> <form METHOD="POST"> <tr> <td>obj2</td> <td>this is obj2</td> <td><input type="hidden" name="id" value="<br /><b>Notice</b>: Undefined index: id in <b>C:\xampp\htdocs\test2\test.php</b> on line <b>30</b><br />"> <input type="submit" value="delete"/></td> </tr> </form> <form METHOD="POST"> <tr> <td>obj3</td> <td>this is obj3</td> <td><input type="hidden" name="id" value="<br /><b>Notice</b>: Undefined index: id in <b>C:\xampp\htdocs\test2\test.php</b> on line <b>30</b><br />"> <input type="submit" value="delete"/></td> </tr> </form> </table> <div><a href="http://localhost/test2/submit.php">点此添加信息</a></div> 1Array() </body></html>
你点击 delete 按钮了吗？
贴出浏览器中的 html
你点击 delete 按钮了吗？
贴出浏览器中的 html
中了hidden的招了
30行的value定错了
真是太感谢了 谢谢帮助！
你点击 delete 按钮了吗？
贴出浏览器中的 html
之前的错误只是之一啊
我数据库里写的是objid $row提的是id 这里错了
改了之后value得到值了 但是为啥post还是空啊
<html> <head> <title>Print test</title> <link href="css.css" rel="stylesheet" type="text/css" /> </head> <body> <br> 显示： <br> <table id="table1"> <tr class="table_head"> <td>objname</td> <td>objinfo</td> </tr> <form METHOD="POST"> <tr> <td>obj1</td> <td>this is obj1</td> <td><input type="hidden" name="id" value="1"> <input type="submit" value="delete"/></td> </tr> </form> <form METHOD="POST"> <tr> <td>obj2</td> <td>this is obj2</td> <td><input type="hidden" name="id" value="2"> <input type="submit" value="delete"/></td> </tr> </form> <form METHOD="POST"> <tr> <td>obj3</td> <td>this is obj3</td> <td><input type="hidden" name="id" value="3"> <input type="submit" value="delete"/></td> </tr> </form> </table> <div><a href="http://localhost/test2/submit.php">点此添加信息</a></div> 1Array() </body></html>
你是指
1Array
(
)
这个是空的吗？
这里要提交（点击 delete 按钮）后才会有值！
你点击 delete 按钮了吗？
贴出浏览器中的 html
之前的错误只是之一啊
我数据库里写的是objid $row提的是id 这里错了
改了之后value得到值了 但是为啥post还是空啊
<html> <head> <title>Print test</title> <link href="css.css" rel="stylesheet" type="text/css" /> </head> <body> <br> 显示： <br> <table id="table1"> <tr class="table_head"> <td>objname</td> <td>objinfo</td> </tr> <form METHOD="POST"> <tr> <td>obj1</td> <td>this is obj1</td> <td><input type="hidden" name="id" value="1"> <input type="submit" value="delete"/></td> </tr> </form> <form METHOD="POST"> <tr> <td>obj2</td> <td>this is obj2</td> <td><input type="hidden" name="id" value="2"> <input type="submit" value="delete"/></td> </tr> </form> <form METHOD="POST"> <tr> <td>obj3</td> <td>this is obj3</td> <td><input type="hidden" name="id" value="3"> <input type="submit" value="delete"/></td> </tr> </form> </table> <div><a href="http://localhost/test2/submit.php">点此添加信息</a></div> 1Array() </body></html>
额、、、 你while还是放form里面去吧。。。
你是指
1Array
(
)
这个是空的吗？
这里要提交（点击 delete 按钮）后才会有值！
点了delete之后发现post有值了
万分感谢！
你点击 delete 按钮了吗？
贴出浏览器中的 html
之前的错误只是之一啊
我数据库里写的是objid $row提的是id 这里错了
改了之后value得到值了 但是为啥post还是空啊
<html> <head> <title>Print test</title> <link href="css.css" rel="stylesheet" type="text/css" /> </head> <body> <br> 显示： <br> <table id="table1"> <tr class="table_head"> <td>objname</td> <td>objinfo</td> </tr> <form METHOD="POST"> <tr> <td>obj1</td> <td>this is obj1</td> <td><input type="hidden" name="id" value="1"> <input type="submit" value="delete"/></td> </tr> </form> <form METHOD="POST"> <tr> <td>obj2</td> <td>this is obj2</td> <td><input type="hidden" name="id" value="2"> <input type="submit" value="delete"/></td> </tr> </form> <form METHOD="POST"> <tr> <td>obj3</td> <td>this is obj3</td> <td><input type="hidden" name="id" value="3"> <input type="submit" value="delete"/></td> </tr> </form> </table> <div><a href="http://localhost/test2/submit.php">点此添加信息</a></div> 1Array() </body></html>
额、、、 你while还是放form里面去吧。。。
还是放里面才对吗？
你点击 delete 按钮了吗？
贴出浏览器中的 html
之前的错误只是之一啊
我数据库里写的是objid $row提的是id 这里错了
改了之后value得到值了 但是为啥post还是空啊
<html> <head> <title>Print test</title> <link href="css.css" rel="stylesheet" type="text/css" /> </head> <body> <br> 显示： <br> <table id="table1"> <tr class="table_head"> <td>objname</td> <td>objinfo</td> </tr> <form METHOD="POST"> <tr> <td>obj1</td> <td>this is obj1</td> <td><input type="hidden" name="id" value="1"> <input type="submit" value="delete"/></td> </tr> </form> <form METHOD="POST"> <tr> <td>obj2</td> <td>this is obj2</td> <td><input type="hidden" name="id" value="2"> <input type="submit" value="delete"/></td> </tr> </form> <form METHOD="POST"> <tr> <td>obj3</td> <td>this is obj3</td> <td><input type="hidden" name="id" value="3"> <input type="submit" value="delete"/></td> </tr> </form> </table> <div><a href="http://localhost/test2/submit.php">点此添加信息</a></div> 1Array() </body></html>
额、、、 你while还是放form里面去吧。。。
还是放里面才对吗？
这么多form 看着不别扭。。。
你可以改为使用复选框来让用户选中数据，复选框的名字为lst[]，所有的复选框都叫这个，然后值为$row[id]，然后通过一个按钮来提交数据到删除界面，php接收lst就能获取到所有的选中数据
而且提交到当前页面的话表单action可以为空
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