这个php页面是用来查看用户的，我在mysql users表中有一个页面，我还有另一个页面要添加用户，但是view\u users.php不起作用，它给了我“警告：mysql\u error（）期望参数1是资源，c:\xampp\htdocs\myproject\view\u users.php第71行中给出null“当我从add\u user.php调用view\u user.php时不起作用2）如果我希望用户只查看不修改数据库，我该怎么办？
<!--for delete Record -->
<?php
$msg="";
$opr="";
if(isset($_GET['opr']))
$opr=$_GET['opr'];
if(isset($_GET['rs_id']))
$id=$_GET['rs_id'];
if($opr=="del")
{
$del_sql=mysql_query("DELETE FROM users_tbl WHERE u_id=$id");
if($del_sql)
$msg="1 Record Deleted... !";
else
$msg="Could not Delete :".mysql_error();
}
echo $msg;
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<link rel="stylesheet" type="text/css" href="css/style_view.css" />
<title>::. Build Bright University .::</title>
</head>
<body>
<div id="style_div" >
<form method="post">
<table width="755">
<tr>
<td width="190px" style="font-size:18px; color:#006; text-indent:30px;">View Users</td>
<td><a href="?tag=susers_entry"><input type="button" title="Add new student" name="butAdd" value="Add New" id="button-search" /></a></td>
<td><input type="text" name="searchtxt" title="Enter name for search " class="search" autocomplete="off"/></td>
<td style="float:right"><input type="submit" name="btnsearch" value="Search" id="button-search" title="Search Users" /></td>
</tr>
</table>
</form>
</div><!--- end of style_div -->
<br />
<div id="content-input">
<form method="post">
<table border="1" width="805px" align="center" cellpadding="3" class="mytable" cellspacing="0">
<tr>
<th style="text-align: center;">No</th>
<th style="text-align: center;">Users Name</th>
<th style="text-align: center;">Password</th>
<th style="text-align: center;">Type</th>
<th style="text-align: center;">Note</th>
<th style="text-align: center;"colspan="2">Operation</th>
</tr>
<?php
$key="";
if(isset($_POST['searchtxt']))
$key=$_POST['searchtxt'];
if($key !="")
$sql_sel=mysql_query("SElECT * FROM users_tbl WHERE username like '%$key%' ");
else
$sql_sel=mysql_query("SELECT * FROM users_tbl");
$i=0;
while($row=mysql_fetch_array($sql_sel)){
$i++;
$color=($i%2==0)?"lightblue":"white";
?>
<tr bgcolor="<?php echo $color?>">
<td><?php echo $i;?></td>
<td><?php echo $row['username'];?></td>
<td><?php echo $row['password'];?></td>
<td><?php echo $row['type'];?></td>
<td><?php echo $row['note'];?></td>
<td align="center"><a href="?tag=susers_entry&opr=upd&rs_id=<?php echo $row['u_id'];?>" title="Upate"><img style="-webkit-box-shadow: 0px 0px 0px 0px rgba(50, 50, 50, 0.75);-moz-box-shadow: 0px 0px 0px 0px rgba(50, 50, 50, 0.75);box-shadow: 0px 0px 0px 0px rgba(50, 50, 50, 0.75);" src="picture/update.png" height="20" alt="Update" /></a></td>
<td align="center"><a href="?tag=view_users&opr=del&rs_id=<?php echo $row['u_id'];?>" title="Delete"><img style="-webkit-box-shadow: 0px 0px 0px 0px rgba(50, 50, 50, 0.75);-moz-box-shadow: 0px 0px 0px 0px rgba(50, 50, 50, 0.75);box-shadow: 0px 0px 0px 0px rgba(50, 50, 50, 0.75);" src="picture/delete.jpg" height="20" alt="Delete" /></a></td>
</tr>
<?php
}
?>
</table>
</form>
</div><!-- end of content-input -->
</body>
</html>
1条答案
按热度按时间vkc1a9a21#
如果没有传递链接标识符，mysql\u error和mysql\u query将采用mysql\u connect的最后一个链接标识符。如果代码从未实际连接到数据库，那么我认为这就是您将得到的错误。确保已连接到mysql服务器并选择了数据库。看到了吗http://php.net/manual/en/function.mysql-connect.php 以及http://php.net/manual/en/function.mysql-select-db.php