我正在为我的网站创建一个wp插件,并且我的网站上有一个表单,必须在数据库中添加一些价值。我的问题是数据不想放入我的数据库中。
我尝试多次更改编码,然后调试功能(数据通过功能可接收),但实际上我的代码如下:
我的首页:
$date_start = $_POST['startDate'];
$date_end = $_POST['endDate'];
$description = $_POST['description'];
$place = $_POST['lieu'];
$value = $_POST['value'];
$user_id = $_POST['id_user'];
$current_user = wp_get_current_user();
$current_user_id = $current_user->id;
$date_current = date('Y/m/d', time());
frdp_add($user_id, $description, $place, $date_start, $date_end, $value, $date_current, $current_user_id);
我的插件页面:
$table_name = $wpdb->prefix . 'frais_deplacement';
function frdp_add($user_id, $description, $place, $date_start, $date_end, $value, $date_current, $current_user_id) {
global $wpdb;
$wpdb->insert(
$table_name,
array(
'id_user' => $user_id,
'description' => $description,
'place' => $place,
'date_start' => $date_start,
'date_end' => $date_end,
'value' => $value,
'created_at' => $date_current,
'created_by' => $current_user_id
)
);
}
答案 0 :(得分:0)
您的代码应如下所示,您跳过了函数中的参数。
$table_name = $wpdb->prefix . 'frais_deplacement';
function frdp_add($current_user_id, $description, $place, $date_start, $date_end, $value, $date_current, $current_user_id) {
global $wpdb;
$wpdb->insert(
$table_name,
array(
'id_user' => $current_user_id,
'description' => $description,
'place' => $place,
'date_start' => $date_start,
'date_end' => $date_end,
'value' => $value,
'created_at' => $date_current,
'created_by' => $current_user_id,
)
);
}
答案 1 :(得分:0)
好吧,我不能将值放入数据库,因为表名不是通过函数来获得的。为此,我必须像这样将 <!doctype html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>R7 FrontEnd Test</title>
<link href="css/style.css" rel="stylesheet" type="text/css">
</head>
<body>
<!-- let's code :D -->
<div class="container">
<div class="box">
<div class="box-head">
<span class="logo">
<img src="./assets/a-fazenda-logo.png" alt="">
</span>
<span class="title">
<h1>RANKING</h1>
</span>
</div>
<div id="ranking" class="main-box">
</div>
</div>
</div>
<script src="./javascripts/js.js"></script>
</html>
插入我的函数中:
global $table_name;