PHP:Retrieving images from a database query issues

Question

I have read many posts but i cannot find my answer. I am developing a food order/delivery website, which has many food cuisine categories, African, Alcohol, American... Each category is meant to have a different header image. So if the admin creates a new restaurant, when they select the restaurant cuisine, the correct header image will automatically display on the main websites products page for that said restaurant.

I have manually inputted the images into the database already, now i am trying to retrieve the database, my or die statement prints that it is not, but i have no error messages, which is confusing me.

    mysqli_report(MYSQLI_REPORT_INDEX);
    if (isset($_GET['rest_id'])) {

        $Rest = $_GET['rest_id'];

        $get_cat_img = "SELECT Cuisine_category
        FROM Rest_Category,Category_img
        INNER JOIN Rest_Details
        ON Rest_Category.Cat_ID = Rest_Details.Cat_ID
        WHERE Rest_Details.Cat_ID='$Rest'";

        $results = mysqli_query($dbc, $get_cat_img) or die("query is not working");
        $row=mysqli_fetch_array($results) or die ("q not working");
        $img=$row['Category_img'];
        echo $row['Category_img'];

        echo '<img src="'.$img.'" alt="background" style="width:100%;height:300px">';           
    }
    mysqli_close($dbc);

`or die("query is not working")` that doesn't help you here, should your query have failed, `or die(mysqli_error($dbc))` will, as will http://php.net/manual/en/function.error-reporting.php — Funk Forty Niner, May 08 '16 at 14:26
Hiya Ralph, I some how knew you would be lurkin, so I already added that to my answer @Fred-ii- — RiggsFolly, May 08 '16 at 14:27
@RiggsFolly *Ah, well will ya lookah dat!* hehe. Let's see if anything comes of it. Spidey sense huh? — Funk Forty Niner, May 08 '16 at 14:28
@Monroe do a var_dump to see what's going through there or not. What's going on right now? Does anything echo at all, look at your HTML source also. also what is the column type for the image and the data? — Funk Forty Niner, May 08 '16 at 14:34
@Monroe see this answer http://stackoverflow.com/a/12785796/ your present method is failing you because of it. also http://stackoverflow.com/a/13225760/ that's why you get those `?` marks. — Funk Forty Niner, May 08 '16 at 14:54

RiggsFolly · Accepted Answer · 2016-05-08T14:23:57.997

1

I think you may have just put a column name in the wrong place in your query.

If Category_img is a column name in the Rest_Category table, this is what you want to do

$get_cat_img = "SELECT Cuisine_category,Category_img
                FROM Rest_Category
                  INNER JOIN Rest_Details ON Rest_Category.Cat_ID = Rest_Details.Cat_ID
                WHERE Rest_Details.Cat_ID='$Rest'";

You can also shorten things a bit by using Alias's, it often makes the SQL code easier to read when it get past the very simple query.

$get_cat_img = "SELECT rc.Cuisine_category,rc.Category_img
                FROM Rest_Category rc
                  INNER JOIN Rest_Details rd ON rc.Cat_ID = rd.Cat_ID
                WHERE rd.Cat_ID='$Rest'";

Also modify your error reporting to actually report the real MYSQL error, it is much more usful that any message you can some up with

Like so

$results = mysqli_query($dbc, $get_cat_img);
if ( $result === false ) {
    echo 'query is not working: ' . mysqli_error($dbc);
    exit;
}

edited May 08 '16 at 14:23

answered May 08 '16 at 13:57

RiggsFolly

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I don't think it likes inner join, i just performed to see if a simple query would work select * FROM .... and the query works, weird – Monroe May 08 '16 at 14:12
Difficult to help here as I dont know your database schema – RiggsFolly May 08 '16 at 14:15
Also make sure you have linked these tables correctly `Rest_Category.Cat_ID = Rest_Details.Cat_ID` without that it is not going to work as you expect – RiggsFolly May 08 '16 at 14:26
@RiggsFolly http://php.net/manual/en/function.error-reporting.php also ;-) just in case. – Funk Forty Niner May 08 '16 at 14:30
I was just looking that statement up. One of many holes in my knowlwdge! – RiggsFolly May 08 '16 at 14:32
@RiggsFolly you think they need a `while` loop here? – Funk Forty Niner May 08 '16 at 14:33
@RiggsFollyi think you are right about the ON... i just deleted that and it works to some extent, adding back in the error message appears, i will look into that. I should have mentioned i have error reporting right at the top of my page – Monroe May 08 '16 at 14:35
Ralph, I think `mysqli_report(MYSQLI_REPORT_INDEX);` is just _adding an extra error reporting to say_ You are not using an index in this query @Fred-ii- – RiggsFolly May 08 '16 at 14:39
@RiggsFolly *10-4 Smokey* – Funk Forty Niner May 08 '16 at 14:40
1

Ralph, well he has obviously solved it judging by the acceptance @Fred-ii- With a little help from his friends – RiggsFolly May 08 '16 at 14:40
found the issue, i believe. You have been great help! just now have to work out why i am seeing a ?mark instead of a the actual image. – Monroe May 08 '16 at 14:50
@RiggsFolly [*Right here Smokey...*](http://stackoverflow.com/questions/37100528/phpretrieving-images-from-a-database-query-issues?noredirect=1#comment61744493_37100528) as per the above comment by the OP. Pretty sure that's the clincher here ;-) – Funk Forty Niner May 08 '16 at 14:55
@Monroe Its far easier to store a path to an image on disk in the database than store the actual image in a blob on the database – RiggsFolly May 08 '16 at 14:59
@RiggsFolly oops! i am very new to this. Is that the better way to receive my goal – Monroe May 08 '16 at 15:18
That is the way I would personnaly approach it – RiggsFolly May 08 '16 at 15:19

PHP:Retrieving images from a database query issues

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