Question

我正在尝试创建一个将textarea数据发送到数据库的php表单。数据库表上传包含3个所需的数据项：

的 USER_ID
的类别
内容

HTML：

<form id="myform" action="php/savetext.php" method="POST" enctype="multipart/form-data">

  <p><strong>Add your story here.</strong></p>

        <input type="hidden" name="fbid" id="fbid">
        <input type="hidden" name="category" id="category">

        <textarea cols="50" rows="10" name="usertext" id="storyArea"></textarea>
        <input type="submit" name="submit" value="Saada" />

        <script type="text/javascript">

            $( "#fbid" ).val( "sdf88d99sd" );
            $( "#category" ).val( "texts" );    

        </script>                       
</form>

savetext.php：

include_once('config.php');

if (isset($_POST['user_id']) && isset($_POST['category']) && isset($_POST['usertext'])) {

    $user_id = mysql_real_escape_string($_POST['fbid']);
    $category = mysql_real_escape_string($_POST['category']);
    $content = mysql_real_escape_string($_POST['usertext']);
    addUser($user_id, $category, $content);
}

else {
    echo 'Upload failed! Try again.';
}

function addUser($user_id, $category, $content) {
    $query = "SELECT id FROM uploads WHERE user_id = '$fbid' LIMIT 1";
    $result = mysql_query($query);
    $rows = mysql_num_rows($result);
    if ($rows > 0) {
        $query = "UPDATE uploads SET user_id = '$user_id', category = '$category' WHERE content = '$content'";
    }
    else {
        $query = "INSERT INTO uploads (user_id, category, content) VALUES ('$user_id', '$category', '$content')";
    }
    mysql_query($query);
    echo 'Upload was succesful. Thank you!';

}

但这不起作用。有关如何纠正此问题的任何想法？

Answer 1

请查看此代码，<form></form>

中没有名称为user_in的字段

if (isset($_POST['fbid']) && isset($_POST['category']) && isset($_POST['usertext'])) {

    $user_id = mysql_real_escape_string($_POST['fbid']);
    $category = mysql_real_escape_string($_POST['category']);
    $content = mysql_real_escape_string($_POST['usertext']);
    addUser($user_id, $category, $content);
}

将textarea数据发送到数据库时遇到麻烦

1 个答案: