怎么将表格中输入的数据反馈到数据库中,以下是有问题的代码:
<form method="post" action="<?php echo $_SERVER['PHP_SELF'];?>"> id: <input type="text" name="id"><br> name: <input type="text" name="name"><br> pass: <input type="text" name="pass"><br> rank: <input type="text" name="rank"><br> <input type="Submit" name="$submit" value=" 提交 ">
<input type="Submit" name="$delete" value=" 删除 ">
(删除时只需输入id。)
//表格
</form> <?php
$db = mysql_connect("localhost", "root","");
mysql_select_db("stu",$db);
if ($id) {
if ($submit) {
$sql = "UPDATE user SET id='$id',name='$name',
pass='$pass',rank='$rank' WHERE id=$id";}}
$result = mysql_query($sql);
?>
//修改<?php
echo "<table border='1'>\n<thead><tr><th>id</th><th>name</th><th>pass</th><th>rank</th></tr></thead>\n</tbody>\n";
$db = mysql_connect("localhost", "root","");
mysql_select_db("stu",$db);
$result = mysql_query("SELECT * FROM user",$db);
if ($myrow = mysql_fetch_array($result)) {
do {
printf("<tr><td>%s</td><td>%s</td><td>%s</td><td>%u</td></tr>\n",
$myrow["id"], $myrow["name"], $myrow["pass"],$myrow["rank"]);
} while ($myrow = mysql_fetch_array($result));
} else {
echo " 对不起,没有找到记录! ";
}
echo "</tbody>\n</table>\n";
?>
</body>
</html>
//输出数据库为 stu,表为user(id varchar(11),name varchar(20),pass varchar(30),rank enum('1','2','3','4')default4,primary key(id));
<form method="post" action="<?php echo $_SERVER['PHP_SELF'];?>"> id: <input type="text" name="id"><br> name: <input type="text" name="name"><br> pass: <input type="text" name="pass"><br> rank: <input type="text" name="rank"><br> <input type="Submit" name="$submit" value=" 提交 ">
<input type="Submit" name="$delete" value=" 删除 ">
(删除时只需输入id。)
//表格
</form> <?php
$db = mysql_connect("localhost", "root","");
mysql_select_db("stu",$db);
if ($id) {
if ($submit) {
$sql = "UPDATE user SET id='$id',name='$name',
pass='$pass',rank='$rank' WHERE id=$id";}}
$result = mysql_query($sql);
?>
//修改<?php
echo "<table border='1'>\n<thead><tr><th>id</th><th>name</th><th>pass</th><th>rank</th></tr></thead>\n</tbody>\n";
$db = mysql_connect("localhost", "root","");
mysql_select_db("stu",$db);
$result = mysql_query("SELECT * FROM user",$db);
if ($myrow = mysql_fetch_array($result)) {
do {
printf("<tr><td>%s</td><td>%s</td><td>%s</td><td>%u</td></tr>\n",
$myrow["id"], $myrow["name"], $myrow["pass"],$myrow["rank"]);
} while ($myrow = mysql_fetch_array($result));
} else {
echo " 对不起,没有找到记录! ";
}
echo "</tbody>\n</table>\n";
?>
</body>
</html>
//输出数据库为 stu,表为user(id varchar(11),name varchar(20),pass varchar(30),rank enum('1','2','3','4')default4,primary key(id));
http://user.qzone.qq.com/447338871/infocenter
http://www.php100.com/html/shipinjiaocheng/PHP100shipinjiaocheng/list_20_8.html
error_reporting(0);
echo "<table border='1'>\n<thead><tr><th>id</th><th>name</th><th>pass</th><th>rank</th></tr></thead>\n</tbody>\n";
$db = mysql_connect("localhost", "root","");
mysql_select_db("stu",$db);
$result = mysql_query("SELECT * FROM user",$db);
if ($myrow = mysql_fetch_array($result)) {
do {$arr[]=$myrow['id'];
printf("<tr><td>%s</td><td>%s</td><td>%s</td><td>%u</td></tr>\n",
$myrow["id"], $myrow["name"], $myrow["pass"],$myrow["rank"]);
} while ($myrow = mysql_fetch_array($result));
} else {
echo " 对不起,没有找到记录! ";
}
echo "</tbody>\n</table>\n";
?>
<form method="post" action=""> id: <input type="text" name="id"><br> name: <input type="text" name="name"><br> pass: <input type="text" name="pass"><br> rank: <input type="text" name="rank"><br> <input type="Submit" name="submit" value=" 提交 ">
<input type="Submit" name="delete" value=" 删除 ">
(删除时只需输入id。)
//表格
</form> <?php
if ($_POST[submit]) {
{
$id=$_POST[id];
$name=$_POST[name];
$pass=$_POST[pass];
$rank=$_POST[rank];
if(!is_null(in_array($id,$arr))){
$sql = "UPDATE user SET id='$id',name='$name',
pass='$pass',rank='$rank' WHERE id='$id'";}
else $sql = "insert into user values('$id','$name','$pass','$rank')";}
}
elseif($_POST[delete]){
$id=$_POST[id];
$sql="delete from user where id='$id'";
}
$result = mysql_query($sql);?> 应该符合你的本意了