Question

I am trying to retrieve the ID of the submitted data that the user enters then display it using jQuery/Ajax. The idea is to display the link of the data or, “quote” page. “Your quote was submitted! Click [link] to go to the page” The link would look like this: http://example.com/quote-123. 123 being the ID of the quote.

This is the part I am using to submit the data using ajax and jQuery.

$(document).delegate("'#submit-quote'", "submit", function(){
        var quoteVal = $(this).find('[name="quote"]').val();
        $.post("add.php", $(this).serialize(), function(data) {
            var like = $('.quote-wrap span iframe');
            $('.inner').prepend('<div class="quote-wrap group">' + like + '<div class="quote"><p>' + quoteVal+ '</p></div></div>');
            // console.log("success");
            var id = parseInt(data);
            alert(id);
        });
        return false;
    });

alert(id) returns NaN.

add.php:

require('inc/connect.php');


    $quote = $_POST['quote'];
    $quotes = mysql_real_escape_string($quote);

    //echo $quotes . "Added to database";

    mysql_query("INSERT INTO entries (quote) VALUES('$quotes')")
    or die(mysql_error());
    echo mysql_insert_id($quote);

I am new to JavaScript and PHP and I’m really trying to learn and understand it. How can I get the id from the submitted quote?

Answer 1

The error in your pastie code says:

Warning: mysql_insert_id() expects parameter 1 to be resource, string given in C:\wamp\www\virallikesproject\add.php on line 12

The resource it is expecting is the link identifier returned by mysql_connect() which I don’t see in your code. Regardless, the mysql_insert_id() behavior if you DONT pass it a resource is to use the last one opened.

Changing this line:

mysql_insert_id($quote);

to this:

mysql_insert_id();

should take care of that error. As far as the overall issue, I’m concerned because I don’t see where you are connecting to your database at all.