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我有一個應用程序，我正在爲幻想足球網站工作。如果你去這裏：如何將數據值傳遞給表單？
http://digitaldemo.net/kickass/a-results.php
你可以看到管理員會看到，當他們去查看/編輯播放器。
UPDATE --------------------------------------
這是
<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>
，這是QB-edit.php的內容：在-results.php呈現到上QB-edit.php形式的數據和應該通數據相關的代碼
<?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>
我收到一條錯誤消息：
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
這是QB-edit.php的第55行：
$result = mysql_query($connect, "SELECT * FROM ff_projections where ID='" . $posted_id . "'") or die ("Error in query");
我去堅果試圖讓這個工作...
你有點迷失了我。我對編程極其新穎...... – Cynthia
在包含這些按鈕的每種表單中，都放置了與每個玩家關聯的記錄的ID號。然後，您可以在qb-edit.php頁面上將該ID作爲'$ _POST'變量獲取，並使用它從數據庫獲取其信息。 –
那麼編輯按鈕表單代碼看起來就像這樣：
？ – Cynthia