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Home/ Questions/Q 7681281
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Editorial Team
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Editorial Team
Asked: May 31, 20262026-05-31T18:18:25+00:00 2026-05-31T18:18:25+00:00

I’m trying to code a tooltip, which uses http://craigsworks.com/projects/simpletip/# simpletip plugin, which returns data

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I’m trying to code a tooltip, which uses http://craigsworks.com/projects/simpletip/# simpletip plugin, which returns data via ajax. However, i can’t seem to display the data from 3 tables, which is fetched via the data image, as each image is assigned to it’s own id. as seen below, i need to display all data from the table, but with 3 tables to read from. in my webpage, a php statement will query data from one table for each category, generated by this tooltips. only a round tooltip came out, not even mysql_error(). i need help asap. this is the only site where i seem to get inputs from others.

<?php

define('INCLUDE_CHECK', 1);
require "../connect.php";

if (!$_POST['img'])
    die(mysql_error());

$img = mysql_real_escape_string(end(explode('/', $_POST['img'])));

$row = mysql_fetch_assoc(mysql_query("SELECT alcoholic.*, non_alcoholic.*,
    non_alcoholic1.* FROM alcoholic INNER JOIN non_alcoholic USING (img) INNER JOIN
    non_alcoholic1 USING (img) WHERE img='" . $img . "' "));

if (!$row)
    die(mysql_error());

echo '<strong>' . $row['name'] . '</strong>

    <p class="descr">' . $row['description'] . '</p>

    <strong>price: $' . $row['price'] . '</strong>
    <small>Drag it to your shopping<br /> cart to purchase it</small>';
?>
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-31T18:18:26+00:00Added an answer on May 31, 2026 at 6:18 pm

    I’m guessing there are three tables with an img column? If so, is this what you need?:

    SELECT
        alcoholic.*, non_alcoholic.*, non_alcoholic1.*
    FROM
        alcoholic
    INNER JOIN non_alcoholic USING (img)
    INNER JOIN non_alcoholic1 USING (img)
    WHERE
        img = '" . $img . "'
    

    Edit: When I said update your OP with the new issue I didn’t mean overwrite it, I meant add it underneath so that this answer (and others if there were any) make sense.

    Anyway, the problem is that $row['name'], etc. don’t refer to anything in the SELECT statement. You need to change alcoholic.*, non_alcoholic.*, non_alcoholic1.* to what it is you’re pulling out, for example:

    SELECT alcoholic.name, alcoholic.description, alcoholic.price ...

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