SQL语句不起作用，我做错了什么> [重复]

I'm creating an admin panel with a form that creates a blog post. When I try to insert this into the database, I'm getting an error. This is my SQL statement:

INSERT INTO blog_posts ('title', 'author', 'tags', 'category', 'body') VALUES ('$title', '$author', '$tags', '$category', '$body');

What am I doing wrong? Full code:

$servername = "host";
$username = "username";
$password = "password";
$dbname = "dbname";

// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
  die("Connection failed: " . mysqli_connect_error());
}

$title = $_POST['title'];
$author = $_POST['author'];
$tags = $_POST['tags'];
$category = $_POST['category'];
$body = $_POST['editor1'];

// sql to insert a record

$sql = "INSERT INTO blog_posts ('title', 'author', 'tags', 'category', 'body') VALUES ('$title', '$author', '$tags', '$category', '$body');";

if (mysqli_query($conn, $sql)) {
  echo "Succes";
} else {
  echo "Error";
}

mysqli_close($conn);

So, what do I do now?

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1. You should not use single quote or double quote for table column field.
2. End with two time semicolon

Instead of:

$sql = "INSERT INTO blog_posts ('title', 'author', 'tags', 'category', 'body') VALUES ('$title', '$author', '$tags', '$category', '$body');";

Try this

 $sql = "INSERT INTO blog_posts (`title`, `author`, `tags`, `category`, `body`) VALUES ('$title', '$author', '$tags', '$category', '$body')";

One thing that is immediately obvious (as stated in comments), is your use of quotes around the field names, quotes are reserved for strings, not field/column names, so where you have;

INSERT INTO blog_posts ('title', 'author', 'tags', 'category', 'body') VALUES 

Should instead be;

INSERT INTO blog_posts (`title`, `author`, `tags`, `category`, `body`) VALUES 

The above uses backticks, you can discard these, but be aware of special words in MySQL

Also to note - you are wide open to SQL injection attacks, make use of prepared statements, these are best, but there are a couple of other options;

With this such as this, as per my comment, it is often appropriate (and helpful) to make use of some of the debugging tools provided to you.
In this instance, mysqli_error would have been of great use to you (and in turn us) as it gives quite a good indication of what went wrong / where