javascript - PHP - Mysql 准备不会执行

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我正在尝试使用名为可编辑网格( http://www.editablegrid.net/en/ )的工具创建 mysql 条目。我刚刚在代码上添加了 2 个文本字段..但 mysql 准备代码似乎不喜欢我正在做的事情。

这就是 php 代码:

    require_once('config.php');         

// Database connection                                   
$mysqli = mysqli_init();
$mysqli->options(MYSQLI_OPT_CONNECT_TIMEOUT, 5);
$mysqli->real_connect($config['db_host'],$config['db_user'],$config['db_password'],$config['db_name']); 

// Get all parameter provided by the javascript
$name = $mysqli->real_escape_string(strip_tags($_POST['name']));
$firstname = $mysqli->real_escape_string(strip_tags($_POST['firstname']));
$uid = $mysqli->real_escape_string(strip_tags($_POST['uid']));
$show = $mysqli->real_escape_string(strip_tags($_POST['show']));
$tablename = $mysqli->real_escape_string(strip_tags($_POST['tablename']));

$return=false;
if ( $stmt = $mysqli->prepare("INSERT INTO ".$tablename."  (name, firstname, uid, show) VALUES (  ?, ?, ?, ? )")) {

    $stmt->bind_param("ssss", $name, $firstname, $uid, $show);
    $return = $stmt->execute();
    $stmt->close();
}             
$mysqli->close();        

echo $return ? "ok" : "error";

这里是 html 代码:

 <div id="addform" style="height:210px;">

            <div class="row">
                <input type="text" id="name" name="name" placeholder="name" />
            </div>

             <div class="row">
                <input type="text" id="firstname" name="firstname" placeholder="firstname" />
            </div>
             <div class="row">
                 <input type="text" id="uid" name="uid" placeholder="uid" />
            </div>

            <div class="row">
                <input type="text" id="show" name="show" placeholder="show" /> 
                </div>

            <div class="row tright">
              <a id="addbutton" class="button green" ><i class="fa fa-save"></i> Apply</a>
              <a id="cancelbutton" class="button delete">Cancel</a>
            </div>
        </div>

它们都通过 JavaScript 连接在一起，并且成功了。我刚刚添加了 2 个文本字段。 Php代码是否有错误。我不允许在整个准备语句中使用 4 个值吗？

Answer 1

代码看起来正确，但变量表名未定义，因此向 html 添加更多文本字段

<div class="row">
<input type="text" id="tablename" name="tablename" placeholder="input table name" />
</div>

或者在php中设置表名

$tablename = 'table name here';