jquery ajax to php script not saving data

Question

I have the following jQuery ajax call to a php script:

actualHtml = $('div').html(); // could this line be causing an issue?
$.ajax({
        type: 'POST',
        url: 'save-html-css-action.php',
        data: { 
           'htmlTextToSave': htmlTextToSave,
           'actualHtml': actualHtml,
           'userId':userId
        },
        success: function(msg){
            alert(msg);
        }
});

php:

$htmlCssToSave = $_POST['htmlTextToSave'];
$userId = $_POST['userId'];
$actualHtml = $_POST['actualHtml'];


$mysqli = new mysqli($servername, $sqlusername, $sqlpassword, $dbname);

/* check connection */
if (mysqli_connect_errno()) {
    //printf("Connect failed: %s\n", mysqli_connect_error());
     echo "Connection failed: ".mysqli_connect_error();
    exit();
}


    $mysqli->query("INSERT INTO user_saved_data (user_html_css_code, dd_id, actual_html) values ('".$htmlCssToSave."',".$userId.",'".$actualHtml."')");
    echo "success";

    /* close connection */
    $mysqli->close();

but when I check the database, the data isn't there. Am I doing something wrong in the jquery/php combo (meaning the ajax call)? I'm getting a javascript "success" alert, so it's hitting the script, but I'm not sure why the info isn't being inserted.

The table datatypes are medium text for both the htmlcsstosave and the actualhtml columns, and int for userid (not the primary key, this is a foreign key to another table)

so I added a an error alert and this is the output

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'margin-0'>
<head> </head>
<body cl' at line 1

Go ahead and find out why, if(!$mysqli->query(....your query)){ echo $mysqli->error; } Also, you're accepting user input and should be using prepared statements to help fight sql injection. — Ohgodwhy
– Ohgodwhy, Commented Mar 26, 2015 at 23:24
.. this is gonna be fun, I'm trying to save html, and its saying bad syntax at somewhere, will update the question as I'm not sure what exactly the syntax it doesn't like is — Abdul Ahmad
– Abdul Ahmad, Commented Mar 26, 2015 at 23:28
No need. Simply change to prepared statements and you'll be good. — Ohgodwhy
– Ohgodwhy, Commented Mar 26, 2015 at 23:29
No need to go digging; I've provided you with the prepared solution below. — Ohgodwhy
– Ohgodwhy, Commented Mar 26, 2015 at 23:36

Ohgodwhy · Accepted Answer · 2015-03-27 04:38:49Z

3

As we discussed, the issue is with escaping and sanitizing data. If you used prepared statements, it will handle it for you. let's have a look at how that would work:

Prepare the statement:

$stmt = $mysqli->prepare("INSERT INTO user_saved_data (user_html_css_code, dd_id, actual_html) values (?,?,?)");

Bind your parameters:

$stmt->bind_param('sis', $htmlCssToSave, $userId, $actualHtml);

Then execute your statement:

$stmt->execute();

Then you should be good to go. The prepared statement should handle the data sanitization for you now.

Resouces:

edited Mar 27, 2015 at 4:38

answered Mar 26, 2015 at 23:36

Ohgodwhy

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3 Comments

Abdul Ahmad Over a year ago

I'm getting an error Call to undefined method mysqli::bind_param()

Abdul Ahmad Over a year ago

actually I think I fixed it, had to set the prepare to another variable $statement and call bind on that

Ohgodwhy Over a year ago

@AbdulAhmad You're totally right. I omitted that in my initial answer. I've updated to reflect that fact; sorry for that!

Aaron Ebejer · Accepted Answer · 2015-03-26 23:29:00Z

2

Can you post the CREATE TABLE statement for the user_saved_data table?

In general, I'd recommend just tracing it through and see where the data gets lost.

For example, can you echo the query you are generating and run in manually in MySQL? You may have a syntax error in the SQL being generated...

edited Mar 26, 2015 at 23:29

answered Mar 26, 2015 at 23:24

Aaron Ebejer

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1 Comment

Ohgodwhy Over a year ago

I understand that as you have a low reputation, you're not able to post comments, but this is for a reason. Please refrain from posting comments as answers. Once you have enough karma (50 or so), you can post comments. I won't downvote you, so that you don't have to climb the hill back, but please delete this.

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