php 使用 WordPress $wpdb 将数据插入到 WordPress 数据库中的表中
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Inserting data into a table in WordPress database using WordPress $wpdb
提问by AmmyKami83
I am starting out plugin development and have followed the tutorials on the WordPress Codex sites. I am now stuck - I have a database called "wp_imlisteningto", where the wp_was inserted using:
我正在开始插件开发并遵循 WordPress Codex 网站上的教程。我现在卡住了 - 我有一个名为“wp_imlisteningto”的数据库，其中wp_插入了使用：
$table_name = $wpdb->prefix . "imlisteningto";
When the plugin is activated.
当插件被激活时。
The database itself has three columns, set up when the plugin is activated:
数据库本身有三列，在插件激活时设置：
$sql = "CREATE TABLE $table_name (
id mediumint(9) AUTO_INCREMENT,
album VARCHAR(50),
artist VARCHAR(50),
PRIMARY KEY (id)
);";
I am trying to insert data (by creating a new row) into this database from a php form.
我正在尝试从 php 表单将数据（通过创建新行）插入到这个数据库中。
Within the WordPress admin, I create a new page which has the very simple form:
在 WordPress 管理员中，我创建了一个具有非常简单形式的新页面：
<form action="/wp-content/plugins/listeningto/formhtml.php" method="post">
Album: <input type="text" name="album" />
Artist: <input type="text" name="artist" />
<input type="submit">
</form>
Which as you can see calls formhtml.php, which is:
正如你所看到的 call formhtml.php，它是：
<?php
global $wpdb;
$wpdb->insert( $table_name, array( 'album' => $_POST['album'], 'artist' => $_POST['artist'] ), array( '$s', '$s' ) );
?>
When I submit the form, I get an Error 500.0when running the plugin in Worpdress on IIS7.0, and a "Page Not Found"when running on another web server which runs apache.
当我提交表单，我得到一个Error 500.0在Worpdress运行插件时IIS7.0，和"Page Not Found"另一台Web服务器运行在运行时apache。
If I change formhtml.phpto:
如果我formhtml.php改为：
<?php
echo $_POST['album'];
echo $_POST['artist'];
?>
Works fine - I get the album and artist that I put in the form. Obviously something I'm doing wrong when inserting the data (in a new row) into the database.
工作正常 - 我得到了我放在表格中的专辑和艺术家。显然，在将数据（在新行中）插入数据库时我做错了。
Any thoughts as to what that might be?
关于那可能是什么的任何想法？
UPDATE
更新
Ok, so if I update formhtml.phpwith this:
好的，所以如果我formhtml.php用这个更新：
<?php
require_once('../../../wp-config.php');
$table_name = $wpdb->prefix . "imlisteningto";
$wpdb->insert( $table_name, array( 'album' => $_POST['album'], 'artist' => $_POST['artist'] ), array( '$s', '$s' ) );
?>
I no longer get an error message, but data still doesn't get put into the database.
我不再收到错误消息，但数据仍然没有放入数据库。
UPDATE 2
更新 2
This worked for me:
这对我有用：
<?php
require_once('../../../wp-config.php');
global $wpdb;
$table_name = $wpdb->prefix . "imlisteningto";
$wpdb->insert( $table_name, array( 'album' => $_POST['album'], 'artist' => $_POST['artist'] ) );
?>
as did this:
就像这样做：
<?php
require_once('../../../wp-load.php');
global $wpdb;
$table_name = $wpdb->prefix . "imlisteningto";
$wpdb->insert( $table_name, array( 'album' => $_POST['album'], 'artist' => $_POST['artist'] ) );
?>
So, for some reason $wpdbwas not working unless I required either wp-configor wp-load.php. If include wp-load.php, $wpdbgets values and all is well.
因此，出于某种原因$wpdb，除非我需要wp-config或wp-load.php。如果包含wp-load.php，则$wpdb获取值并且一切正常。
回答by Pramod
including
包含
require_once('../../../wp-config.php');
worked for me
为我工作
回答by Ajay Patel
Try this..
尝试这个..
<?php
global $wpdb;
$wpdb->insert( $table_name, array( 'album' => "$_POST['album']", 'artist' => "$_POST['artist']" ) );
?>
Ex :
前任 ：
<?php
global $wpdb;
$wpdb->insert($table_name , array('chart_name' => "Line Chart" ,'chart_type' => "trends",'status' => 0));
?>
回答by ultrageek
You've probably figured this out by now, but no one addressed it here. Your sample code has '$s' in the 3rd parameter (2nd array), but that should be '%s' because it's for value-formatting. The WP Codex says [http://codex.wordpress.org/Class_Reference/wpdb]that this format parameter for $wpdb->insert() is optional.
你现在可能已经明白了这一点，但没有人在这里解决它。您的示例代码在第三个参数（第二个数组）中有 '$s'，但这应该是 '%s'，因为它用于值格式化。WP Codex 说 [ http://codex.wordpress.org/Class_Reference/wpdb]$wpdb->insert() 的这个格式参数是可选的。
回答by Lodi Cronje
I think there are 2 mistakes in you sql string.
我认为您的 sql 字符串中有 2 个错误。
Think it should be the $table_namevariable should be concatenated
认为应该是$table_name变量应该被连接
$sql = "CREATE TABLE" . $table_name . "(
id mediumint(9) AUTO_INCREMENT,
album VARCHAR(50),
artist VARCHAR(50),
PRIMARY KEY (id)
)";
and remove ;on the last line.
并删除;最后一行。
