我有一个申请 我为一个幻想足球场工作。
您可以看到当管理员去查看/编辑玩家时会看到什么。
更新 - - - - - - - - -
这是关于结果.php的相关代码,可以提供数据, 应该 将数据传送到qb-edit.php的窗体:
<table cellspacing="0" cellpadding="5" border="1" width="560">
<tr style="text-align:center">
<td style="text-align:left ; width:175px">Player Name</td>
<td>Team</td>
<td>Pass Yds</td>
<td>Pass TDs</td>
<td>Int Thrown</td>
<td>Rush Yds</td>
<td>Rush TDs</td>
<td>Overall Pts</td>
<td>TFP</td>
</tr>
<?php
$result = mysql_query("SELECT ID, Player, Team, Pass_Yds, Pass_TDs, Int_Thrown, Rush_Yds, Rush_TDs, Overall_Pts, Total_Fantasy_Pts FROM ff_projections WHERE Position = QB ORDER BY Pass_Yds DESC;");
while($row = mysql_fetch_array($result))
{
echo "<tr style="text-align:center"><td style="text-align:left">{$row[ Player ]}</td>";
echo "<td>{$row[ Team ]}</td>";
echo "<td>{$row[ Pass_Yds ]}</td>";
echo "<td>{$row[ Pass_TDs ]}</td>";
echo "<td>{$row[ Int_Thrown ]}</td>";
echo "<td>{$row[ Rush_Yds ]}</td>";
echo "<td>{$row[ Rush_TDs ]}</td>";
echo "<td>{$row[ Overall_Pts ]}</td>";
echo "<td>{$row[ Total_Fantasy_Pts ]}</td>";
echo "<td><form action="qb-edit.php" method="post"><input type="hidden" name="ID" value="". $row[ ID ] .""><input type="submit" name="submit" value="Edit"></form></td></tr>";
}
?>
</table>
这是qb- edit. php 的内容 :
<?php
$posted_id = $_POST[ ID ];
?>
<!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" />
<title>Untitled Document</title>
<style>
body { font-family:Arial, Helvetica, sans-serif ;
font-size:14px ;
}
.form { width:350px ;
margin-auto ;
}
label { clear:both ;
display:block ;
float:left ;
padding-right:8px ;
line-height:26px ;
}
input[type=text] { float:right ;
width:163px ;
height:18px ;
margin:3px 0px ;
}
input[type=text].short { width:30px ;
margin-right:132px ;
}
input[type=submit] { clear:both ;
float:left ;
margin-top:20px ;
margin-bottom:20px ;
}
select { float:right ;
margin-right:118px ;
}
</style>
</head>
<body>
<div class="form">
<?php
echo $posted_id;
$result = mysql_query($connect, "SELECT * FROM ff_projections where ID= " . $posted_id . " ") or die ("Error in query");
if ($row = mysql_fetch_array($result)) {
echo "<form method= post action= add_player.php >";
echo "<label for= Player >Player Name:</label> <input type= text name= Player value= " . $row[ Player ] . " />";
echo "<label for= Pass_Yds >Pass Yds:</label> <input class= short type= text name= Pass_Yds value= " . $row[ Pass_Yds ] . " />";
echo "<label for= Pass_TDs >Pass TDs:</label> <input class= short type= text name= Pass_TDs value= " . $row[ Pass_TDs ] . " />";
echo "<label for= Int_Thrown >Int Thrown:</label> <input class= short type= text name= Int_Thrown value= " . $row[ Int_Thrown ] . " />";
echo "<label for= Rush_Yds >Rush Yds:</label> <input class= short type= text name= Rush_Yds value= " . $row[ Rush_Yds ] . " />";
echo "<label for= Rush_TDs >Rush TDs:</label> <input class= short type= text name= Rush_TDs value= " . $row[ Rush_TDs ] . " />";
echo "<label for= Overall_Pts >Overall Pts:</label> <input class= short type= text name= Overall_Pts value= " . $row[ Overall_Pts ] . " />";
echo "<input type= submit name= submit value= Update Player />";
echo "<input type= hidden name= id value= " . $row[ ID ] . " />";
echo "</form>";
}
?>
</div>
</body>
</html>
我有一个错误说:
PHP Warning: mysql_query() expects parameter 2 to be resource, string given in C:\xampp\htdocs\kickass\qb-edit.php on line 55, referer: http://localhost/kickass/a-results.php
这是qb- edit. php 的55行 :
$result = mysql_query($connect, "SELECT * FROM ff_projections where ID= " . $posted_id . " ") or die ("Error in query");
我快疯了,想把事情办成...