数据库名为practice,表名为product.
要把把product_name里的值赋给product_name_f。
要如何实现。我写了一个不行,而且有错误,请指正:代码如下:
<form name="form1" method="post" action="">
<input type="submit" name="Submit" value="add">
</form>
<?
$dbhost = 'localhost';//mysql的数据库名
$dbname = 'practice';//mysql用户
$dbuser = 'root';
$dbpw = '';
$db=new database;
$db->connect($dbhost, $dbuser, $dbpw, $dbname);//把product_name里的值赋给product_name_f
$product_name_f=$db->query("select * from product,id asc");
$data=$db->fetch_array($product_name_f);if($action=="add")
{
$operate_database="insert into product(product_name_f[id])values ('$data[id]')";
if ($db->query($operate_database))
{
echo $var_option_succ;
echo 'alert"HEEEO"';
exit;
}
}
?>
要把把product_name里的值赋给product_name_f。
要如何实现。我写了一个不行,而且有错误,请指正:代码如下:
<form name="form1" method="post" action="">
<input type="submit" name="Submit" value="add">
</form>
<?
$dbhost = 'localhost';//mysql的数据库名
$dbname = 'practice';//mysql用户
$dbuser = 'root';
$dbpw = '';
$db=new database;
$db->connect($dbhost, $dbuser, $dbpw, $dbname);//把product_name里的值赋给product_name_f
$product_name_f=$db->query("select * from product,id asc");
$data=$db->fetch_array($product_name_f);if($action=="add")
{
$operate_database="insert into product(product_name_f[id])values ('$data[id]')";
if ($db->query($operate_database))
{
echo $var_option_succ;
echo 'alert"HEEEO"';
exit;
}
}
?>
$db->connect($dbhost, $dbuser, $dbpw, $dbname);去掉$db=new database;
提示:
Fatal error: Call to a member function on a non-object in E:\PHP\data-in.php on line 13
<input type="submit" name="Submit" value="add">
</form>
<?
$conn=mysql_pconnect("localhost","root","");
mysql_select_db('practice',$conn) or die("无法连接到数据库,请与管理员联系!");//把product_name里的值赋给product_name_f
$product_name_f=$db->query("select * from product,id asc");
$data=$db->fetch_array($product_name_f);if($action=="add")
{
$operate_database="insert into product(product_name_f[id])values ('$data[id]')";
if ($db->query($operate_database))
{
echo $var_option_succ;
echo 'alert"HEEEO"';
exit;
}
}
?>改成这样后的错误提示:
Fatal error: Call to a member function on a non-object in E:\PHP\data-in.php on line 9不知道这行哪里错了
$product_name_f=$db->query("select * from product,id asc");
sql是
update tab_name set col_1=col_2 where where_condi;