有一个product_db的数据库,其下有个productb,下面是该表的结构。+----------+-------------+------+-----+---------+-------+
| Field | Type | Null | Key | Default | Extra |
+----------+-------------+------+-----+---------+-------+
| id | int(10) | YES | | NULL | |
| name | varchar(20) | YES | | NULL | |
| price | float | YES | | NULL | |
| num | int(100) | YES | | NULL | |
| add_time | datetime | YES | | NULL | |
+----------+-------------+------+-----+---------+-------+添加的表单是pad.html大致如下
<html><body>
<center >产品表添加 </center>
<form action="pad1.php" method="POST">
<p>
<p>
<table align=center>
<tr align=center>
<td align=center>
编号:<input type="text" name="id">
<p>
品名:<input type="text" name="name">
<p>
价格:<input type="text" name="price">
<p>
数量:<input type="text" name="num">
<p>
日期:<input type="text" name="add_time">
<p>
<center>
<input type="submit" value="进行添加"></a>
</center>
</form>
<p>
<body></html>
提交的服务器端是pad.php大致如下<center>
<a href="pad1.html">继续添加</a><center>
<?php
$link = mysql_connect("localhost","root","123456"); //用户名和密码依据本机具体设置
$eset= mysql_query("set names gbk");
$sdb = mysql_select_db("product_db",$link); //数据库依据本机具体设置
$id=$_POST['id'];
$name = $_POST['name'];
$pirce=$_POST['pirce'];
$num=$_POST['num'];
$add_time=$_POST['add_time'];
$ins = "INSERT INTO product_db.productb (id,name,pirce,num,add_time) VALUES ('$id','$name','$pirce','$num','$add_time');"; //表依据本机具体设置
$eins =mysql_query($ins);
$query = "select * from productb;";
$result = mysql_db_query("product_db", $query);
while ($r = mysql_fetch_array($result))
{
$id = $r["id"];
$name=$r["name"];
$pirce=$r["pirce"];
$num=$r["num"];
$add_time=$r["add_time"];
echo " $id";
echo " $name";
echo " $pirce";
echo " $num ";
echo "$add_time ";
}
mysql_free_result($result);
?> 当都添加好,提交后,显示的结果用浏览器的刷新就自动重复添加上次提交的内容了,重复刷新重复提交显示!如何做到浏览器的刷新后不进行提交处理? (麻烦大家提供个范例,谢谢了)
| Field | Type | Null | Key | Default | Extra |
+----------+-------------+------+-----+---------+-------+
| id | int(10) | YES | | NULL | |
| name | varchar(20) | YES | | NULL | |
| price | float | YES | | NULL | |
| num | int(100) | YES | | NULL | |
| add_time | datetime | YES | | NULL | |
+----------+-------------+------+-----+---------+-------+添加的表单是pad.html大致如下
<html><body>
<center >产品表添加 </center>
<form action="pad1.php" method="POST">
<p>
<p>
<table align=center>
<tr align=center>
<td align=center>
编号:<input type="text" name="id">
<p>
品名:<input type="text" name="name">
<p>
价格:<input type="text" name="price">
<p>
数量:<input type="text" name="num">
<p>
日期:<input type="text" name="add_time">
<p>
<center>
<input type="submit" value="进行添加"></a>
</center>
</form>
<p>
<body></html>
提交的服务器端是pad.php大致如下<center>
<a href="pad1.html">继续添加</a><center>
<?php
$link = mysql_connect("localhost","root","123456"); //用户名和密码依据本机具体设置
$eset= mysql_query("set names gbk");
$sdb = mysql_select_db("product_db",$link); //数据库依据本机具体设置
$id=$_POST['id'];
$name = $_POST['name'];
$pirce=$_POST['pirce'];
$num=$_POST['num'];
$add_time=$_POST['add_time'];
$ins = "INSERT INTO product_db.productb (id,name,pirce,num,add_time) VALUES ('$id','$name','$pirce','$num','$add_time');"; //表依据本机具体设置
$eins =mysql_query($ins);
$query = "select * from productb;";
$result = mysql_db_query("product_db", $query);
while ($r = mysql_fetch_array($result))
{
$id = $r["id"];
$name=$r["name"];
$pirce=$r["pirce"];
$num=$r["num"];
$add_time=$r["add_time"];
echo " $id";
echo " $name";
echo " $pirce";
echo " $num ";
echo "$add_time ";
}
mysql_free_result($result);
?> 当都添加好,提交后,显示的结果用浏览器的刷新就自动重复添加上次提交的内容了,重复刷新重复提交显示!如何做到浏览器的刷新后不进行提交处理? (麻烦大家提供个范例,谢谢了)
解决方案 »
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