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- 6
- 78567 2013-03-14
格式将数据插入数据库
我正在编写一个简单的插件,该插件在数据库中创建一个名称为"newsletter"的表,并提供一个短代码以在页面中放置注册表单. 表单包含"名称"和"电子邮件". 我在数据库中插入表格数据(名称+电子邮件)时遇到问题. 我写了这个:

<?php $name = $_POST['name']; $email = $_POST['email']; function insertuser(){ global $wpdb; $table_name = $wpdb->prefix . "newsletter"; $wpdb->insert($table_name, array('name' => $name, 'email' => $email) ); } ?>

,但ID不起作用.我应该怎么做才能从表单中获取数据并将其插入表中?
I am writing a simple plugin that create a table with name "newsletter" in database and provide a shortcode to put a registration form in pages. the form contain "name" and "email". i have problem with inserting the form data(name+email) in database. i wrote this:

<?php $name = $_POST['name']; $email = $_POST['email']; function insertuser(){ global $wpdb; $table_name = $wpdb->prefix . "newsletter"; $wpdb->insert($table_name, array('name' => $name, 'email' => $email) ); } ?>

but id does not work. what should i do to get data from form and insert into table?
database forms table

- 您是否在新闻通讯表中添加了任何前缀?
  
  Have you added any prefix to newsletter table?
  - 0
  - 2013-03-14
  - Vinod Dalvi
- 到底在哪里添加前缀?数据库中的表具有前缀
  
  adding prefix where exactly? the table in database have prefix
  - 0
  - 2013-03-14
  - pixelweb
- 在表名之前,因为您已经使用了$table_name=$ wpdb->prefix."通讯";在您的代码中,在表名新闻稿之前添加了wordpress前缀,因此,如果您没有向表名添加任何前缀,则只能使用像这样的表名$table_name="newsletter";
  
  before table name because you have used this $table_name = $wpdb->prefix . "newsletter"; in your code which adds wordpress prefix before the table name newsletter so if you have not added any prefix to table name than only use table name like this $table_name = "newsletter";
  - 0
  - 2013-03-14
  - Vinod Dalvi
- 函数内部的两个变量`name`和`email`是** unknown **.您必须在函数内部定义它们,或者如果在其他地方需要它们,则将它们声明为"global"(函数的_outside_和_inside_).
  
  The two variables `name` and `email` are **unknown** inside the function. You have to either define them inside the function, or if they are needed elsewhere, declare them `global` (both _outside_ and _inside_ the function).
  - 0
  - 2013-03-14
  - tfrommen
- @VinodDalvi:我在数据库中添加了表的前缀.
  
  @VinodDalvi : i added th eprefix for table in database.
  - 0
  - 2013-03-14
  - pixelweb
- 我定义了内部函数的名称和电子邮件,但没有任何反应.
  
  i defined the name and email inside function but nothing happen.
  - 0
  - 2013-03-14
  - pixelweb
- 您添加到表中的前缀是什么?告诉我带有前缀的完整表名.
  
  What is the prefix you have added to the table? tell me the full table name with prefix.
  - 0
  - 2013-03-15
  - Vinod Dalvi

1 个回答

投票数

Mike Madern · Accepted Answer · 2013-03-14

6

2013-03-14

函数内部两个变量$name和$email是未知的.您必须通过将global $wpdb更改为global $wpdb, $name, $email:

使它们在全局范围内可用

 require_once('../../../wp-load.php');

/**
 * After t f's comment about putting global before the variable.
 * Not necessary (http://php.net/manual/en/language.variables.scope.php)
 */
global $name = $_POST['name'];
global $email = $_POST['email'];

function insertuser(){
  global $wpdb, $name, $email;
  $table_name = $wpdb->prefix . "newsletter";
  $wpdb->insert($table_name, array('name' => $name, 'email' => $email) ); 
}

insertuser();

或者,您可以将变量放在函数的参数中:

 require_once('../../../wp-load.php');

$name = $_POST['name'];
$email = $_POST['email']

function insertuser( $name, $email ) {
  global $wpdb;

  $table_name = $wpdb->prefix . 'newsletter';
  $wpdb->insert( $table_name, array(
    'name' => $name,
    'email' => $email
  ) );
}

insertuser( $name, $email );

或者,没有功能:

 require_once('../../../wp-load.php');

global $wpdb;

$name = $_POST['name'];
$email = $_POST['email'];
$table_name = $wpdb->prefix . "newsletter";
$wpdb->insert( $table_name, array(
    'name' => $name,
    'email' => $email
) );

The two variables $name and $email are unknown inside the function. You have to make them globally available inside it by changing global $wpdb into global $wpdb, $name, $email:

require_once('../../../wp-load.php');

/**
 * After t f's comment about putting global before the variable.
 * Not necessary (http://php.net/manual/en/language.variables.scope.php)
 */
global $name = $_POST['name'];
global $email = $_POST['email'];

function insertuser(){
  global $wpdb, $name, $email;
  $table_name = $wpdb->prefix . "newsletter";
  $wpdb->insert($table_name, array('name' => $name, 'email' => $email) ); 
}

insertuser();

Or, you can put the variables in the function's arguments:

require_once('../../../wp-load.php');

$name = $_POST['name'];
$email = $_POST['email']

function insertuser( $name, $email ) {
  global $wpdb;

  $table_name = $wpdb->prefix . 'newsletter';
  $wpdb->insert( $table_name, array(
    'name' => $name,
    'email' => $email
  ) );
}

insertuser( $name, $email );

Or, without function:

require_once('../../../wp-load.php');

global $wpdb;

$name = $_POST['name'];
$email = $_POST['email'];
$table_name = $wpdb->prefix . "newsletter";
$wpdb->insert( $table_name, array(
    'name' => $name,
    'email' => $email
) );

那就是我写的;)关于何时撰写评论以及何时足以回答问题,我仍然有疑问.;)变量仍然必须声明为`global` _outside_函数.

That's what I wrote. ;) I'm still having issues with when to write a comment and when it's enough for an answer. ;) The variables still have to be declared `global` _outside_ the function, though.
- 0
- 2013-03-14
- tfrommen
哈哈,是的,我在发布答案后看到了您的评论:-)我的评论/回答规则是,如果OP必须在代码中更改多个规则,请务必回答;-)我将在变量$name和$email

Haha, yes I saw your comment after I posted my answer :-) my rule of commenting/answering is, if OP has to change more than one rule in the code, always answer ;-) I'll add `global` to the variables `$name` and `$email`
- 1
- 2013-03-14
- Mike Madern
嗯,好吧,您似乎对范围是正确的(因为在这种情况下,函数_outside_是全局范围,而不是类).但是,如果将变量声明为全局变量(现在决定要做什么),则必须先声明,然后(在下一行或分号后)设置一个值.

Ah, okay, you seem to be right with the scope (because in this case, _outside_ the function **is** the global scope, and not a class). However, if you declare the variables global (what you decided to do now), you first have to declare, and then (in the next line, or after a semicolon) set a value.
- 0
- 2013-03-14
- tfrommen
当用户单击提交表单时,表单的动作引用该文件中的文件调用:regiostration-form.php,我现在具有以下代码: `前缀."通讯"; $ wpdb->insert($table_name,array('name'=> $name,'email'=> $email));;复制代码 } ?>` 但它不再起作用.哪里不对了?

when user click on submit form the form's action refer to an file call :regiostration-form.php in this file i have this code now: `prefix . "newsletter"; $wpdb->insert($table_name, array('name' => $name, 'email' => $email) ); } ?> ` but it does not work again. anything wrong?
- 0
- 2013-03-14
- pixelweb
声明函数后,您是否要调用`insertuser()`函数?

You do call the `insertuser()` function after you declare the function?
- 0
- 2013-03-14
- Mike Madern
@MikeMadern我是否必须在函数后编写:'insertuser()'?

@MikeMadern do i have to write: 'insertuser()' after function?
- 0
- 2013-03-14
- pixelweb
定义函数不会自动执行它.您必须调用该函数才能执行.看到我的答案中的代码;-)

defining a function doesn't automatically executes it. You have to call the function in order to execute. See the code in my answer ;-)
- 0
- 2013-03-14
- Mike Madern
我使用了您的代码,但遇到两个错误:注意:"试图在第8行中获取非对象的属性"和"致命错误:在第9行中的非对象上调用成员函数insert()"

i used your code but i got two errors: Notice: `Trying to get property of non-object in line 8` and `Fatal error: Call to a member function insert() on a non-object in line 9`
- 0
- 2013-03-14
- pixelweb
这不是WordPress加载的文件吗?在脚本顶部`require``wp-load.php`文件以加载WordPress库.

It isn't a file loaded by WordPress right? `require` the `wp-load.php` file on top of your script to load the WordPress library.
- 0
- 2013-03-14
- Mike Madern
我使用此代码并正常工作:`前缀."通讯"; $ wpdb->insert($table_name,array('name'=> $name,'email'=> $email)));复制代码 ?> `特别感谢所有人Mike Madern

i used this code and worked fine: `prefix . "newsletter"; $wpdb->insert( $table_name, array( 'name' => $name, 'email' => $email ) ); ?> ` Thanks all guys specially Mike Madern
- 0
- 2013-03-14
- pixelweb
是的,请用完整答案更新答案,然后我会接受.真诚的

yes, please update the answer with complete answer then i will accept it. Sincererly
- 0
- 2013-03-14
- pixelweb
我为您更新了答案；)

I updated my answer for you ;)
- 0
- 2013-03-15
- Mike Madern