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php添加到数据库表,该怎么解决

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Release: 2016-06-13 12:01:51
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php添加到数据库表
自学php,也看了一些解决的方法,但还是不行。遇到以下情况,不能成功向添加数据到数据库里,希望大神指点。这是我的表

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"><br /><html xmlns="http://www.w3.org/1999/xhtml"><br /><head><br /><meta http-equiv="Content-Type" content="text/html; charset=utf-8" /><br /><title>无标题文档</title><br /></head><br /><body><br /><form method="POST"><br /><table><br /><tr><td>c_id</td><td><input type="c_id" size="20"></td></tr><br /><tr><td>c_name</td><td><input type="c_name" size="20"></td></tr><br /><tr><td>sector</td><td><input type="sector" size="50"></td></tr><br /><tr><td>e_address</td><td><input type="e_address" size="50"></td></tr><br /><tr><td>mail_address</td><td><input type="mail_address" size="50"></td></tr><br /><tr><td>classification</td><td><input type="classification" size="50"></td></tr><br /><tr><td>started_date</td><td><input type="started_date" size="50"></td></tr><br /><tr><td>contact</td><td><input type="contact" size="50"></td></tr><br /><tr><td>con_address</td><td><input type="con_address" size="50"></td></tr><br /><tr><td><input type="Submit" value="添加"></td></tr><br /></table><br /></form><br /><?php<br />$db = mysql_connect("localhost", "root", "root");<br />mysql_select_db("test", $db);<br />if(isset($_POST['submit'])&&$_POST['submit'])<br />{<br />     $sql="insert into customers_id(c_id,c_name,sector,e_address,mail_address,classification,started_date,contact,con_address) <br />     values('$_POST[c_id]','$_POST[c_name]','$_POST[sector]','$_POST[e_address]','$_POST[mail_address]','$_POST[classification]''$_POST[started_date]','$_POST[contact]','$_POST[con_address]''now()'";<br />     mysql_query($sql);<br />echo "成功";<br />}<br />?><br />echo "注册成功";<br /><br />?><br /><br /><br /><br /></body><br /></html>
Copy after login

------解决方案--------------------
1、表单控件的定义不对,都改成下面这样的:
   

2、提交按钮没有name值,不会进入if分支里面。
name="submit">

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