我的MySQL查询有什么问题？

Question

So, I have a form that posts to my php file using ajax, and succeeds. But the following query doesn't insert anything. Can someone help me understand what I'm doing wrong?

My php file:

<?php

include 'connect.php' ;

$type = mysql_real_escape_string($_POST['type']);

$title = mysql_real_escape_string($_POST['title']);

$content = mysql_real_escape_string($_POST['content']);

if ($type == 'Just Text') {

    mysql_query("INSERT INTO articles (title, type, thisisaninteger, content) VALUES ('".$title."', '".$type."', 0, '".$content."')")or die("MySQL Error: " . mysql_error());

}

?>

My connect.php:

<?php

$dbhost = "localhost"; 
$dbname = "example";
$dbuser = "test"; 
$dbpass = "test"; 

mysql_connect($dbhost, $dbuser, $dbpass) or die("MySQL Error: " . mysql_error());  
mysql_select_db($dbname) or die("MySQL Error: " . mysql_error());

?>  

Answer 1

If you aren't receiving any errors and the INSERT just doesn't happen, it is most likely because the if statement fails to be true. Verify that $type actually matches Just Text .

You should also be inserting values using prepared statements , and use PDO or MySQLi - this article will help you decide which.

Answer 2

first, echo "something" after the if statement and recall the data with your ajax post. you can find out if your if statement is working, then try formatting your variables like so

mysql_query("INSERT INTO articles (title, type, thisisaninteger, content) VALUES ('$title', '$type', 0, '$content')")or die("MySQL Error: " . mysql_error());

Answer 3

I just want to throw in an official vote/recommendation in favor of switching to a parameterized SQL statement , too. In spite of the use of mysql_real_escape_string, schlepping a SQL statement together via string concatenation is neither necessary nor a good idea. Honestly, I find a prepared statement much, much easier to read than the typical string-concatenation exercise, as well:

$stmt = $dbh->prepare("SELECT * FROM users WHERE USERNAME = ? AND PASSWORD = ?");
$stmt->execute(array($username, $password));

Answer 4

Alright, it was a stupid mistake on my side. There were columns I didn't include and they were not being assigned a value. Thanks everyone for helping out.