Question

我有捐赠页面，当用户点击捐赠时，它会将数据发布到名为test.php的php文件中我正在尝试这样做我的第一个尝试回显名字和姓氏但是这不起作用最终我想要这个php页面运行MySQL查询来更新数据库中的total_Donation行，这里是我的主要php页面。

位于文件顶部的数据库代码

<?php
$con = mysql_connect("localhost","root","null");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}

mysql_select_db("snr", $con);

$names_sql = "SELECT first_Name, last_Name FROM donate WHERE user_ID > 0";
$names_query = mysql_query($names_sql)or die(mysql_error());
$rsNames= mysql_fetch_assoc($names_query);

if(isset($_POST['donation']) && $_POST['donation'] != '')
{
$donation = mysql_real_escape_string($_GET['donation']);
$fname = mysql_real_escape_string($_GET['first_Name']);
$lname = mysql_real_escape_string($_GET['last_Name']);
$donate_sql = "UPDATE `donate` SET donate_Total = donate_Total + '{$donation}' WHERE      first_Name = '{$fname}' AND last_Name = '{$lname}'";
}
mysql_close($con);
?>

这是html的表单部分

 form method ="post" action="test.php">
<table>

    <tr><td><label>Runner:</label></td>
    <td>
    <select>
    <?php do{?>
    <option> <?php echo $rsNames['first_Name'];?> <?php echo $rsNames['last_Name'];?></option>
    <?php } while ( $rsNames= mysql_fetch_assoc($names_query))?>
    </select>
    </td>
    </tr>
    <tr><td><label>Donation £</label></td><td><input type="text" maxlength="9"    value="0.00" name="donation"/></td></tr>
    <tr><td><input id="submit" type="submit" value="DONATE"/></td></tr>

</table>
</form>

该选项在用户点击捐赠时获取所有名字和姓氏我希望它运行$ donation_sql但是我得到的是错误说明了不明的索引，我甚至在test.php中尝试以下只需回显first_Name就会产生同样的错误。

<?php

echo $_POST['first_Name'];

?>

有人可以帮助我，谢谢。

Answer 1

你需要给select选择一个name属性：

<select name="first_Name">
<?php while ( $rsNames= mysql_fetch_assoc($names_query)):?>
 <option value="<?php echo htmlspecialchars($rsNames['first_Name']).' '.htmlspecialchars($rsNames['last_Name']);?>"> [option displayed to the user here]</option>
<?php endwhile;?>
</select>

当然使用$ _POST数组，而不是$ _GET，因为你正在使用POST方法。

Answer 2

\确保为select设置名称，并且在选项标记

中有valua attr

<select name="first_Name">

<otpion value="<?php echo $rsNames['first_Name'];?>"><?php echo $rsNames['first_Name'];?> 
        <?php echo $rsNames['last_Name'];?>
</option>

</select>

Answer 3

的index.php

<?php
$con = mysql_connect("localhost","root","null");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}

mysql_select_db("snr", $con);

$names_sql = "SELECT first_Name, last_Name FROM donate WHERE user_ID > 0";
$names_query = mysql_query($names_sql)or die(mysql_error());
?>
<form method ="post" action="test.php">
<table>

    <tr><td><label>Runner:</label></td>
    <td>
    <select name="name">
    <?php
    while($list = mysql_fetch_array($names_query))
    {
    ?>
    <option value="<?php echo $list['first_Name'] . ' ' . $list['last_Name']; ?>">
        <?php echo $list['first_Name'] . ' ' . $list['last_Name']; ?>
    </option>
    <?php
    }
    ?>
    </select>
    </td>
    </tr>
    <tr><td><label>Donation £</label></td><td><input type="text" maxlength="9" value="0.00" name="donation" /></td></tr>
    <tr><td><input id="submit" type="submit" name="send" value="DONATE"/></td></tr>

</table>
</form>

test.php的

<?php
$con = mysql_connect("localhost","root","null");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}

mysql_select_db("snr", $con);

if(isset($_POST['donation']) && $_POST['donation'] != '')
{
$names = explode(' ',$_POST['name']);
$first_name= $names[0];
$last_name= $names[1];
$donation = mysql_real_escape_string($_POST['donation']);
$fname = mysql_real_escape_string($first_name);
$lname = mysql_real_escape_string($last_name);

$donate_sql = "UPDATE `donate` SET donate_Total = donate_Total + '" .$donation. "' WHERE  first_Name = '" .$fname. "' AND last_Name = '" .$lname. "'";

echo 'DEBUG (remove after OK): <br>' .$donate_sql. '<br>';
$res = mysql_query($donate_sql);

echo 'Thanks ' .$first_name. ' ' .$last_name. '<br>';

}
mysql_close($con);
?>

就是这样！

关于php post操作没有运行的建议

3 个答案: