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我正在尝试创建一个表单以允许用户将表单中的数据更新为数据库中的现有数量。这是我迄今为止所拥有的，它的价值似乎翻了一番。我在想我需要从数据库中提取值，然后从表单中添加数据。

<?php 
$username = "username";
$password = "password";
$hostname = "localhost";

//connection to the database
$dbhandle = mysql_connect($hostname, $username, $password)
 or die("Unable to connect to MySQL");
echo "<font face=tahoma color=#ff000><b>Connected to MySQL</b></font><br><br>";

//select a database to work with
$selected = mysql_select_db("pdogclan_points",$dbhandle)
  or die("Did this change");

// Formulate Query
$_POST["filter"];
$memid = mysql_real_escape_string($_POST["Member_ID"]);
$query = sprintf("SELECT Member_ID, Bank, Reward_1, Reward_2, Reward_3 FROM Points_Rewards WHERE Member_ID = '$memid'") or die("Could Not Formulate the Query");

//execute the SQL query and return records
$result = mysql_query($query);

// Check result
// This shows the actual query sent to MySQL, and the error. Useful for debugging.
if (!$result) {
    $message  = 'Invalid query: ' . mysql_error() . "\n";
   $message .= 'Whole query: ' . $query;
    die($message);
}

//fetch tha data from the database
while ($row = mysql_fetch_array($result))   
echo "<table width=750 cellspacing=2 cellpadding=2 border=2>
        <tr>
            <td bgcolor=#000000 width=150><font face=tahoma color=white>ID: {$row['Member_ID']}</font></td>". 
            "<td width=150><font face=tahoma>Bank: {$row['Bank']}</td>". 
            "<td width=150><font face=tahoma>Reward 1: {$row['Reward_1']}</td>". 
            "<td width=150><font face=tahoma>Reward 2: {$row['Reward_2']}</td> ". 
            "<td width=150><font face=tahoma>Reward 3: {$row['Reward_3']}</td>
        </tr>
  </table><br></font>";//display the results
// Formulate Update Query
$_POST["submit"];
$memid = mysql_real_escape_string($_POST["Member_ID"]);
$query = sprintf("SELECT Member_ID, Bank, Reward_1, Reward_2, Reward_3 FROM Points_Rewards WHERE Member_ID = '$memid'") or die("Could Not Formulate the Query");

while ($row = mysql_fetch_array($result))
{
$bankdb = $row['Bank'];
$reward1db = $row['Reward_1'];
$reward2db = $row['Reward_2'];
$reward3db = $row['Reward_3'];
}
echo $bank;
echo $reward1;
echo $reward2;
echo $reward3;
$memid = mysql_real_escape_string($_POST["Member_ID"]);
$bank = $_POST['bank'];
$reward1 = $_POST['reward1'];
$reward2 = $_POST['reward2'];
$reward3 = $_POST['reward3'];



$query = "UPDATE Points_Rewards Set Bank = ('$bank' + '$bankdb'), Reward_1 =    ('$reward1' + '$reward1'), Reward_2 = ('$reward2' + '$reward2'), Reward_3 = ('$reward3' + '$reward3') WHERE Member_ID = '$memid'";
$result = mysql_query($query) or die(mysql_error());



if(mysql_query($query)){
echo "updated";}
else{
echo "fail";}

//close the connection
mysql_close($dbhandle);
?>
4

3 回答 3

1

只需使用基本 HTML 创建一个表单，将您从数据库中获取的数据存储在 PHP 变量中，然后使用 PHP 标签显示该数据，如下所示：

<form action="..." method="post" >
<?php
$memid = mysql_real_escape_string($_POST["Member_ID"]);
$query = sprintf("SELECT Member_ID, Bank, Reward_1, Reward_2, Reward_3 FROM Points_Rewards WHERE Member_ID = '$memid'") or die("Could Not Formulate the Query");

while ($row = mysql_fetch_array($result))
{
?>
<input type="text" name="r1" value="<?php echo $row['Reward_1']; ?>" /> ;
<input type="text" name="r2" value="<?php echo $row['Reward_2']; ?>" /> ;
<input type="text" name="r3" value="<?php echo $row['Reward_3']; ?>" /> ;
...    
<?php 
}
?>   

...
</form>
于 2013-04-04T23:12:34.950 回答
0

这是结构

 // if a form is submitted
 if(isset($_POST['submit'])) {
 $memid = $_POST["Member_ID"];

 //SELECT or INSERT or UPDATE your DATABASE. Yes use PDO and prepared statements.
$query = $dbh->prepare("SELECT Member_ID, Bank, Reward_1, Reward_2, Reward_3 FROM Points_Rewards WHERE Member_ID = '$memid'")

//don't forget to bind parameters
$sth->bindParam(':memid', $memid, PDO::PARAM_INT);
$sth->execute(...);

//the loop
while ($row = $sth->fetch(PDO::FETCH_ASSOC)) {
echo '';
}

//close the if statement
}

//write the form
<form method="post"/>
<input name="Member_ID" type="text" required/>
<input name="submit" type="submit" value="submit" />
</form>
于 2013-04-04T23:24:05.307 回答
0

您可以在 SQL 中的表值上使用运算符 - 它看起来像这样：

$query = "UPDATE Points_Rewards Set Bank = (Bank + '$bankdb'), Reward_1 =    (Reward_1 + '$reward1'), Reward_2 = (Reward_2 + '$reward2'), Reward_3 = (Reward_3 + '$reward3') WHERE Member_ID = '$memid'";
于 2013-04-04T23:10:38.890 回答