wp_insert_post custom post type

1 Views

I’m inserting custom post type with following code: 

if(isset($_POST['submit'])){

    $post_id = wp_insert_post( array(
                    'post_status' => 'publish',
                    'post_type' => 'custom_type',
                    'post_title' => 'Some post',
                    'post_content' => 'Lorem ipsum'
                ) );

}

and I’m getting error page with; Title: “WordPress failure notice” and text in page “Are you sure you want to do this?

Read More

Please try again.”

But when I try to insert post_type: post then is everything ok.
And everything is ok when i try to insert without isset; 

$post_id = wp_insert_post( array(
                        'post_status' => 'publish',
                        'post_type' => 'custom_type',
                        'post_title' => 'Some post',
                        'post_content' => 'Lorem ipsum'
                    ) );

Anybody had same problem?

Post Views: 1

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4 comments

  1. I have not found solution except this (AND IT WORKS):

    if(isset($_POST['submit'])){

    $post_id = wp_insert_post( array(
                    'post_status' => 'publish',
                    'post_type' => 'post',
                    'post_title' => 'Some post',
                    'post_content' => 'Lorem ipsum'
                ) );

$post_type = 'custom_type';

$query = "UPDATE {$wpdb->prefix}posts SET post_type='".$post_type."' WHERE id='".$post_id."' LIMIT 1";

GLOBAL $wpdb; 

$wpdb->query($query);

}

  3. I got mine to work with this:

    PHP, placed before the header:

    if(isset($_POST['new_post']) == '1') {

$new_post = array(
      'ID' => '',
      'post_type' => 'customtype', // Custom Post Type Slug
      'post_status' => 'publish',
      'post_title' => $_POST['post_title'],
    );

$post_id = wp_insert_post($new_post);
$post = get_post($post_id);
}

    and in the HTML: 

    <form method="post" action="">

      <input name="post_title" type="text" />

      <input type="hidden" name="new_post" value="1" />
      <input type="submit" name="submit" value="Post" />

</form>

  4. Try this out this is the form header

    <form id="insert_new_lesson" name="insert_new_lesson" method="post" action="" class="lesson-form">;

    The if statement I make is this

    if ( 'POST' == $_SERVER['REQUEST_METHOD'] && !empty( $_POST['action'] ) && 'insert_new_lesson' == $_POST['action'] )