问题
I am using a jQuery AJAX request to a page called like.php
that connects to my database and inserts a row. This is the like.php
code:
<?php
// Some config stuff
define(DB_HOST, 'localhost');
define(DB_USER, 'root');
define(DB_PASS, '');
define(DB_NAME, 'quicklike');
$link = mysql_connect(DB_HOST, DB_USER, DB_PASS) or die('ERROR: ' . mysql_error());
$sel = mysql_select_db(DB_NAME, $link) or die('ERROR: ' . mysql_error());
$likeMsg = mysql_real_escape_string(trim($_POST['likeMsg']));
$timeStamp = time();
if(empty($likeMsg))
die('ERROR: Message is empty');
$sql = "INSERT INTO `likes` (like_message, timestamp)
VALUES ('$likeMsg', $timeStamp)";
$result = mysql_query($sql, $link) or die('ERROR: ' . mysql_error());
echo mysql_insert_id();
mysql_close($link);
?>
The problematic line is $likeMsg = mysql_real_escape_string(trim($_POST['likeMsg']));
. It seems to just return an empty string, and in my database under the like_message
column all I see is blank entries. If I remove mysql_real_escape_string()
though, it works fine.
Here's my jQuery code if it helps.
$('#like').bind('keydown', function(e) {
if(e.keyCode == 13) {
var likeMessage = $('#changer p').html();
if(likeMessage) {
$.ajax({
cache: false,
url: 'like.php',
type: 'POST',
data: { likeMsg: likeMessage },
success: function(data) {
$('#like').unbind();
writeLikeButton(data);
}
});
} else {
$('#button_container').html('');
}
}
});
All this jQuery code works fine, I've tested it myself independently.
Any help is greatly appreciated, thanks.
回答1:
Are you 1000% sure that $_POST["likeMsg"]
actually contains something?
As for mysql_real_escape_string()
returning an empty value, the manual says there is only one situation where that can happen:
Note: A MySQL connection is required before using mysql_real_escape_string() otherwise an error of level E_WARNING is generated, and FALSE is returned. If link_identifier isn't defined, the last MySQL connection is used.
this doesn't seem to be the case here though, as you do have a connection open. Strange.
回答2:
As the other answers don't make clear what exactly to do, here's my:
When you do
$db_connection = new mysqli($SERVER, $USERNAME, $PASSWORD, $DATABASE);
you need to escape like this:
$newEscapedString = $db_connection->real_escape_string($unescapedString);
NOTE: Because people are downvoting this (WTF!?), here's the official page of the official php manual that says EXACTLY what i have posted: real_escape_string @ PHP Manual.
回答3:
For people who might be finding this again now, I just ran into this problem as I'm migrating from PHP5 to PHP7. I'm changing from
string mysql_real_escape_string(string $unescaped, [resource $link = NULL])
to:
string mysqli_real_escape_string(mysqli $link, string $escapestr)
So, in other words, the database $link is no longer optional and moves to the first argument position. If left out, it returns an empty string, without an error, apparently.
回答4:
Do a var_dump of $_POST['likeMsg'], and a var_dump of $likeMsg. That gives you information on what goes wrong.
回答5:
mysql_real_escape_string() will return blank response if you have not made connection to database ...
来源:https://stackoverflow.com/questions/3005135/mysql-real-escape-string-just-makes-an-empty-string