Вам не хватает запятой.
'{$post_category}' post_title =
'{$post_category}' , post_title =
Вот система в PHP, которая содержит информацию о категориях из базы данных. Я реализовал удаление, но запрос обновления не работает, может ли кто-нибудь помочь мне разобраться в ошибке?
код ниже:
<?php // ob_start(); ?>
<?php
if (isset($_GET['update'])) {
$post_id = $_GET['update'];
$query = "select * from posts where post_id = {$post_id}";
$select_categories = mysqli_query($connection, $query);
while ($row = mysqli_fetch_assoc($select_categories)) {
$post_id = $row['post_id'];
$post_category_id = $row['post_category_id'];
$post_title = $row['post_title'];
$post_author = $row['post_author'];
$post_date = $row['post_date'];
$post_image = $row['post_image'];
$post_content = $row['post_content'];
$post_tags = $row['post_tag'];
$post_comment_count = $row['post_comment_count'];
$post_status = $row['post_status'];
}
}
?>
<?php
if (isset($_POST['subs'])) {
$post_authors = $_POST['post_author'];
$post_statuss = $_POST['post_status'];
$post_tagss = $_POST['post_tags'];
$post_contents = $_POST['post_content'];
$post_titles = $_POST['post_title'];
$post_category = $_POST['post_category'];
$post_images = $_FILES['image']['name'];
$post_image_temp = $_FILES['image']['tmp_name'];
move_uploaded_file($post_image_temp, "../image/$post_image");
$queries = "update posts set post_category_id = '{$post_category}' post_title = '{$post_titles}', post_author = '{$post_authors}', post_date = now(), post_image = '{$post_image}', post_content = '{$post_contents}', post_tag='{$post_tags}', post_comment_count='{$post_comment_count}', post_status = '{$post_status}' where post_id = {$post_id}";
$update_post = mysqli_query($connection, $queries);
if (!$update_post)
echo "Stupid";
}
?>
<form action="posts.php" method="post" enctype="multipart/form-data">
<div class="form-group">
<label for="title">Post title</label>
<input value="<?php if (isset($post_title)){ echo $post_title;} ?>" type="text" class="form-control" name="post_title">
</div>
<div class="form-group">
<label for="title">Post Category Id</label>
<br>
<select name="post_category" id=""> <!-- post_category is the one that takes the value -->
<?php
if (isset($_GET['update'])) {
$cat_ids = $_GET['update'];
$query = "select * from categories";
$select_categories = mysqli_query($connection, $query);
confirm($select_categories);
while ($row = mysqli_fetch_assoc($select_categories)) {
$cat_id = $row['cat_id'];
$cat_title = $row['cat_title'];
echo "<option value='{$cat_id}'>{$cat_title}</option>";
}
}
?>
</select>
</div>
<div class="form-group">
<label for="title">Post Author</label>
<input value="<?php if (isset($post_author)){ echo $post_author;} ?>" type="text" class="form-control" name="post_author">
</div>
<div class="form-group">
<label for="title">Post Status</label>
<input value="<?php if (isset($post_status)){ echo $post_status;} ?>" type="text" class="form-control" name="post_status">
</div>
<div class="form-group">
<label for="title">Post Image</label>
<img width="100" src="../images/<?php echo $post_image; ?>" alt="">
<input type="file" name="image">
</div>
<div class="form-group">
<label for="title">Post Tags</label>
<input value="<?php if (isset($post_tags)){ echo $post_tags;} ?>" type="text" class="form-control" name="post_tags">
</div>
<div class="form-group">
<label for="title">Post Content</label>
<textarea class="form-control" name="post_content" id="" cols="30" rows="10"><?php if (isset($post_content)){ echo $post_content;} ?></textarea>
</div>
<div class="form-group">
<input class="btn btn-primary" type="submit" name="subs" value="Update">
</div>
</form>
Работы: 1. Удаление ссылок одним кликом
Форма будет извлечена из БД, и все значения будут показаны в определенном поле
Пользователь может обновить
Значения будут возвращены в db, как только будет нажата кнопка «Обновить».
Пожалуйста, помогите мне обновить ..
php,web,web-deployment,web-deployment-project,