Ajax & PHP - Fetching Mysql table with Ajax not being printed

Question

I am working on a commenting system. I need to be able to post (Insert into database) and the post automatically has to be visible on the site as well.

I have two PHP files, one which is the main page (index.php) and one to connect to the database and fetch (connect_to_database.php).

From connect_to_database.php I connect to the database, and try to fetch row. Then Ajax script is supposed to print into a div the information.

I am able to post, and I can check that the information is added to the database, but I do not see any of the database rows printed.

Could the problem be with the Ajax script?

Thanks for your help,

connect_to_database.php

...
$connect=mysqli_connect($mysql_host, $mysql_user, $mysql_password, $mysql_database);
//$connect=mysqli_connect("localhost", "commenter","phyBq2JT", "udb_commenter" );

// Check connection
if (mysqli_connect_errno())
{
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$db_select = mysqli_select_db($connect,'udb_commenter');
$sql="SELECT * FROM $TABLE_NAME";
$result=mysqli_query( $connect, $sql);


$array = mysqli_fetch_row($result);
echo json_encode($array);
?>

index.php Here is only the part of the Ajax code. Above it, I have a working connection to the database.

<div class="output-post" id="output">
                        <script language="JavaScript" id="print" type="text/javascript" src="jquery.js">
                            $(function ()
                            {
                                $.ajax({

                                    url: 'connect_to_database.php',
                                    data: "",
                                    dataType: 'json',
                                    success: function(data)
                                    {
                                        var id = data[0];
                                        var name = data[1];
                                        var email = data[2];
                                        var type = data[3];
                                        var message = data[4];
                                        var date = data[5];

                                        $('.output-post').html("<b>id: </b>"+id+"<b> name: </b>"+name+email+type+message+date);
                                    }

                                });
                            });

                        </script>
</div>

The sql table:

CREATE TABLE IF NOT EXISTS `posts` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `name` varchar(40) NOT NULL,
  `email` varchar(60) NOT NULL,
  `type` varchar(40) NOT NULL,
  `message` text NOT NULL,
  `date` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
  PRIMARY KEY (`id`)
);

console.log() will write into the javascript console of your browser. you can access this console. i don't know which browser you use but you can try to press F12 to open it up. — Manuel Richarz
– Manuel Richarz, Commented Mar 30, 2014 at 14:50
After the changes made below: SyntaxError: Using //@ to indicate sourceMappingURL pragmas is deprecated. Use //# instead jquery.min.js:1 Error: ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js is being assigned a //# sourceMappingURL, but already has one SyntaxError: missing } after function body example.de:34 — teodr
– teodr, Commented Mar 30, 2014 at 14:55
move your javascript code in an own <script>-block instead of writing it into jquery-script-tag — Manuel Richarz
– Manuel Richarz, Commented Mar 30, 2014 at 15:00

Manuel Richarz · Accepted Answer · 2014-03-30 14:38:52Z

1

i think "data" in your success-method is not one row this are rows. so you have to access one row at once or use a "foreach".

<div class="output-post" id="output">
                        <script language="JavaScript" id="print" type="text/javascript" src="jquery.js">
                            $(function ()
                            {
                                $.ajax({

                                    url: 'connect_to_database.php',
                                    data: "",
                                    dataType: 'json',
                                    success: function(data)
                                    {
                                        jQuery.each(data.rows, function() {
                                           var id = this[0];
                                           var name = this[1];
                                           var email = this[2];
                                           var type = this[3];
                                           var message = this[4];
                                           var date = this[5];

                                           $('.output-post').append("<b>id: </b>"+id+"<b> name: </b>"+name+email+type+message+date);
                                        });
                                    }

                                });
                            });

                        </script>
</div>

also change your php-code into the following:

$array = mysqli_fetch_row($result);
echo "{'rows':" . json_encode($array) . "}";
?>

edited Mar 30, 2014 at 14:38

answered Mar 30, 2014 at 14:07

Manuel Richarz

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7 Comments

teodr Over a year ago

Still nothing is printed. I was using the code for row printing, to see if anything could be printed at all.

Manuel Richarz Over a year ago

can you just call the connect_to_database.php directly in another browser-window and send us the json you'll get. you can remove the data from this json, we just need the structure of it.

teodr Over a year ago

What I get is : ["27","name","[email protected]","new_post","Hello, this is me. I am the creator of Commenter. \r\nPlease feel free to post or comment.","2014-03-30 15:12:34"]

teodr Over a year ago

I was just changing and making sure I did everything as you suggested. The json output is still the same, only the first row and I still do not see anything printed.

Manuel Richarz Over a year ago

ok can you replace "$('.output-post').append(" with "alert(" ?

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