У меня проблема с обновлением контента в базе данных. ошибка:
Query FailedYou have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'post_content = 'kjgljkkjhklj ', post_image = '45739895_2381062595269282_8123898' at line 1
это весь мой код:
if (isset($_GET['p_id'])) {
$the_post_id = $_GET['p_id'];
}
$query = "SELECT * FROM posts WHERE post_id = $the_post_id";
$select_posts_by_id = mysqli_query($connection, $query);
while ($row = mysqli_fetch_assoc($select_posts_by_id)) {
$post_id = $row['post_id'];
$post_author = $row['post_author'];
$post_title = $row['post_title'];
$post_category_id = $row['post_category_id'];
$post_status = $row['post_status'];
$post_image = $row['post_image'];
$post_content = $row['post_content'];
$post_tags = $row['post_tags'];
$post_comment = $row['post_comment_count'];
$post_date = $row['post_date'];
}
// if update post button is clicked
if (isset($_POST['update_post'])) {
$post_author = $_POST['post_author'];
$post_title = $_POST['post_title'];
$post_category_id = $_POST['post_category'];
$post_status = $_POST['post_status'];
$post_image = $_FILES['image']['name'];
$post_image_temp = $_FILES['image']['tmp_name'];
$post_content = $_POST['post_content'];
$post_tags = $_POST['post_tags'];
move_uploaded_file($post_image_temp, "../images/{$post_image}");
if (empty($post_image)) {
$query = "SELECT * FROM posts WHERE post_id = $the_post_id ";
}
$query = "UPDATE posts SET ";
$query .= "post_title = '{$post_title}', ";
$query .= "post_category_id = '{$post_category_id}', ";
$query .= "post_date = now(), ";
$query .= "post_author = '{$post_author}', ";
$query .= "post_status = '{$post_status}', ";
$query .= "post_tags = '{$post_tags}' ";
$query .= "post_content = '{$post_content}', ";
$query .= "post_image = '{$post_image}' ";
$query .= "WHERE post_id = {$the_post_id}"; // this is from the get request
$update_post = mysqli_query($connection, $query);
confirmQuery($update_post);
}
это форма ниже:
<form action = "" method = "POST" enctype = "multipart/form-data">
<div class = "form-group">
<label for = "title">Post Title</label>
<input value = "<?php echo $post_title; ?>" type = "text" name = "post_title" class = "form-control" required = "true">
</div>
<div class = "form-group">
<label for = "post_category">Post Categories</label>
<select name = "post_category" id = "" class = "form-control form-control-md">
<?php
$query = "SELECT * FROM categories";
$select_categories = mysqli_query($connection, $query);
confirmQuery($select_categories); // this is from functions.php
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 = "author">Post Author</label>
<input value = "<?php echo $post_author; ?>" type = "text" name = "post_author" class = "form-control" required = "true">
</div>
<div class = "form-group">
<label for = "post_status">Post Status</label>
<input value = "<?php echo $post_status; ?>" type = "text" name = "post_status" class = "form-control" required = "true">
</div>
<div class = "form-group">
<label for = "image">Post Image</label>
<img width = "100px" src = "../images/<?php echo $post_image; ?>">
<input type = "file" name = "image" required = "true">
</div>
<div class = "form-group">
<label for = "post_tags">Post Tags</label>
<input value = "<?php echo $post_tags; ?>" type = "text" name = "post_tags" class = "form-control" required = "true">
</div>
<div class = "form-group">
<label for = "post_content">Post Content</label>
<textarea name = "post_content" class = "form-control" id = "" cols = "30" rows = "10" required = "true">
<?php echo $post_content; ?>
</textarea>
</div>
<div class = "form-group">
<input type = "submit" name = "update_post" class = "btn btn-primary" value = "Update Post">
</div>
</form>
@Jarzon и да, и нет, это не запятая после post_image, поскольку ГДЕ идет дальше
Это открыто для SQL-инъекций. Параметризовать.
так какого кода не хватает? Можете ли вы указать, какой код следует добавить? Благодарность :)






FWIW, мне легче читать:
$query = "
UPDATE posts
SET post_title = '$post_title'
, post_category_id = '$post_category_id'
, post_date = now()
, post_author = '$post_author'
, post_status = '$post_status'
, post_tags = '$post_tags'
, post_content = '$post_content'
, post_image = '$post_image'
WHERE post_id = $the_post_id;
";
... но также важно заменить эти строки правильно параметризованным запросом
попробуй заменить
$query .= "WHERE post_id = {$the_post_id}";
с участием
$query .= "WHERE post_id = {$the_post_id} ";
Это должно сделать волшебство.
В вашем коде SQL отсутствует , после значений post_tags и post_image.