伙计们,我有这些代码来更新数据库中的记录。
请p ..检查这些是否正确?<?php
session_start();
include "db.php";
$username = $_SESSION['username'];
$query="SELECT * FROM members where username='".mysql_real_escape_string($username)."'";
$result=mysql_query($query);
$num=mysql_numrows($result);
mysql_close();
$userid =mysql_result($result,$i, 'userid');
$firstname =mysql_result($result,$i,'firstname');
$lastname =mysql_result($result,$i,'lastname');
$username =mysql_result($result,$i,'username');
$email =mysql_result($result,$i,'email');
$age =mysql_result($result,$i,'age');
?>
<form action="update.php" method="post">
<input type="hidden" name="u_userid" value="<? echo "$userid" ?>">
<table>
<tr><td>ID:</td> <td><? echo "$userid"?></td></tr>
<tr><td>First Name:</td> <td> <input type="text" name="u_firstname" value="<? echo "$firstname"?>"></td></tr>
<tr><td>Last Name: </td> <td><input type="text" name="u_lastname" value="<? echo "$lastname"?>"></td></tr>
<tr><td>Username:</td> <td> <input type="text" name="u_username" value="<? echo "$username"?>"></td></tr>
<tr><td>Email:</td> <td> <input type="text" name="u_email" value="<? echo "$email"?>"></td></tr>
<tr><td>Age: </td> <td><input type="text" name="u_age" value="<? echo "$age"?>"></td></tr>
<tr><td></td><td><input type="Submit" value="Update Info">
</td></tr>
</form>
<tr><td></td><td><form action="form2.html" method="post">
<input type="submit" value="Cancel"></form></td></tr>
</table
并在update.php中
<?php
session_start();
include "db.php";
$userid =mysql_result($result, 'userid');
$firstname =mysql_result($result, 'firstname');
$lastname =mysql_result($result, 'lastname');
$username =mysql_result($result, 'username');
$email =mysql_result($result, 'email');
$age =mysql_result($result, 'age');
include "db.php";
$query="UPDATE members SET firstname='$u_firstname' , lastname='$u_lastname', username='$u_username' , email='$u_email', age='$u_age' ";
$result=mysql_query($query);
mysql_close();
echo "Record Updated
<br><br>
<form action=\"form2.html\" >
<input type=\"submit\" value=\"ok!\" />
</form>
";
?>
实际上它显示了很多像这样的错误/警告
警告:mysql_result():提供的参数不是第7行的C:\ web \ htdocs \ salome \ this \ update.php中的有效MySQL结果资源
警告:mysql_result():提供的参数不是第8行的C:\ web \ htdocs \ salome \ this \ update.php中的有效MySQL结果资源
警告:mysql_result():提供的参数不是第9行的C:\ web \ htdocs \ salome \ this \ update.php中的有效MySQL结果资源
警告:mysql_result():提供的参数不是第10行的C:\ web \ htdocs \ salome \ this \ update.php中的有效MySQL结果资源
警告:mysql_result():提供的参数不是第11行的C:\ web \ htdocs \ salome \ this \ update.php中的有效MySQL结果资源
警告:mysql_result():提供的参数在第12行的C:\ web \ htdocs \ salome \ this \ update.php中不是有效的MySQL结果资源
我会做什么..
请帮帮我... :-(
答案 0 :(得分:5)
您正在尝试从数据库中获取结果(但即使不执行查询),而您必须通过$ _POST数组
从表单中使用已编辑的数据<?php
session_start();
include "db.php";
$user_id = mysql_real_escape_string($_SESSION['userid']);
$firstname = mysql_real_escape_string($_POST['u_firstname']);
$lastname = mysql_real_escape_string($_POST['u_lastname']);
$lastname = mysql_real_escape_string($_POST['u_lastname']);
$username = mysql_real_escape_string($_POST['u_username']);
$email = mysql_real_escape_string($_POST['u_email']);
$age = mysql_real_escape_string($_POST['u_age']);
//etc
$query = "UPDATE members SET firstname='$firstname', lastname='$lastname',
username='$username' , email='$email', age='$age'
WHERE userid = '$user_id'";
$result = mysql_query($query) or trigger_error(mysql_error().$query);
请注意,用户标识应来自SESSION,而不是FORM
你应该从不编辑用户ID。
答案 1 :(得分:-1)
在更新查询中使用如下,
$query="UPDATE members SET firstname='$u_firstname' , lastname='$u_lastname', username='$u_username' , email='$u_email', age='$u_age' where userid = '$u_userid' ";
在页面末尾使用mysql_close();
(即在提取结果之后)。