数据库添加数据总是提示添加失败

<div class="d1">
	 <span><h1 align="center">注册会员</h1></span>
	 <form action="" method="post">
	 <table class="ta1">
	 	<tr><td align="right">姓名:</td><td><input type="text" name="vip_name"></td></tr>
	 	<tr><td align="right">联系电话:</td><td><input type="text" name="vip_tellphone"></td></tr>
	 	<tr><td align="right">充值时间:</td><td><input type="text" name="vip_registertime"></td></tr>
	 	<tr><td align="right">充值金额:</td><td><input type="text" name="vip_recharge"></td></tr>
	 	<tr><td align="right">住址:</td><td><input type="text" name="vip_address"></td></tr>
	 </table>
	 <div class="span1"><input type="submit" name="add" value="注册">&nbsp;&nbsp;&nbsp;&nbsp;
	 	&nbsp;&nbsp; <input type="reset" name="reset" value="重置"></div>
	 </form>
</div>
<?php
if(isset($_POST["add"]))
{
require_once "config.php";
$name=$_POST["vip_name"];
$tellphone=$_POST["vip_tellphone"];
$registertime=$_POST["vip_registertime"];
$recharge=$_POST["vip_recharge"];
$address=$_POST["vip_address"];
$sql="insert into info_vip(name,tellphone,registertime,recharge,address)
 values('$name','$tellphone','$registertime','$recharge','$address')";
if(mysqli_query($connect,$sql))
  echo "<script>alert('注册成功');this.location.href='select.php'</script>";
else
  echo "<script>alert('注册失败');this.location.href='add.php'</script>";
mysqli_close($connect);
}
?>

 

另一个

<?php

@$connect=mysqli_connect("1.com","root","root");

if(!$connect)

 die( "连接数据库服务器失败".mysqli_connect_error()."<br>");

mysqli_set_charset($connect,"utf8");

$sql_database="create database if not exists vip";

if(mysqli_query($connect,$sql_database))

 echo "数据库创建成功";

else 

 echo "数据库创建失败";

$connect1=mysqli_select_db($connect,"vip");

if(!$connect1)

 die("连接数据库失败".mysqli_error($connect)."<br>");

$sql_table="create table if not exists info_vip(id int not null auto_increment,

name varchar(20) not null,

tellphone varchar(11) not null,

registertime date not null,

recharge int(10) not null,

address varchar(50) not null,

operation varchar(25) not null,

primary key(id)

)";

if(mysqli_query($connect,$sql_table))

 echo "数据表创建成功";

else 

 echo "数据表创建失败";

m

ysqli_close($connect);

?>

表info_vip  operation varchar(25) not null,

 

operation 为非空值,但是你的sql并没有给operation 赋值,所以插入失败了。重新建立info_vip表,去掉operation的 not null。

 

要么sql语句给operation 一个初始值

 

$sql="insert into info_vip(name,tellphone,registertime,recharge,address,operation ) values('$name','$tellphone','$registertime','$recharge','$address','operation初始值')";

 

帮助到你能点个采纳吗,谢谢~

失败肯定是后台处理的有问题啊。你要看你执行sql的地方,是否sql语法有问题等