Displaying images from an array using a function

Question

I am new to PHP and I am trying to print out 5 pictures to which the references are stored inside a string, separated by new line.

Here's what I have so far :

function displayTeam($heroes){  
    $herolist = array();

    $conn = odbc_connect('Dotalyzer', '', '');
    if (!$conn) {
        exit("Connection Failed: " . $conn);
    }
    $sql = "SELECT * FROM HeroStats ORDER BY Heroname";
    $rs = odbc_exec($conn, $sql);
    if (!$rs) {
        exit("Error in SQL");
    }
        foreach(preg_split("/((\r?\n)|(\r\n?))/", $heroes) as $line){
            while (odbc_fetch_row($rs)) {
                $heroname = odbc_result($rs, "HeroName");

                if ($line == $heroname) {

                    $heroimage = odbc_result($rs, "Portrait");
                    $herolist[] = $heroimage . ".png";
                }
            }
        }

    odbc_close($conn);

    echo " <div class='heroimage_wrapper'>";
    for ($i = 0; $i < 5; $i++) {
        echo " <img src='Hero icons/$herolist[$i]' class='heroimage'>";
    }
        echo " </div>";
}

I want to print each picture from the array, the references are stored in an Access database.

I am calling the function which generates my $heroes string using:

 $(document).ready(function(){
        $('.button').click(function(){
            var clickBtnValue = $(this).val();
            var id = $(this).attr('data-unique-id');
            var ajaxurl = 'ajax/ajax',
                data =  {
                    'action': clickBtnValue,
                    'id': id
                };
                console.log(clickBtnValue);
            $.post(ajaxurl, data, function (response) {
                // Response div goes here.
                alert(response.replace('\\n', '\n'));
            });
        });

    });

can you provide some output, what is the error you get, is there a response? what is the content of the header response? — Oussama Elgoumri
– Oussama Elgoumri, Commented Dec 30, 2014 at 21:07
@OussamaELGOUMRI in the alert I get the content of the $heroes string and then it prints out some HTML code followed by Undefined offset from 1 to 4 , for the img src — Bogdan Molinger
– Bogdan Molinger, Commented Dec 30, 2014 at 21:13
what is the name of your php file, and is ajax url pointing to the correct file? — Oussama Elgoumri
– Oussama Elgoumri, Commented Dec 30, 2014 at 21:21

Vitalii · Accepted Answer · 2014-12-30 21:10:36Z

1

You can get 5 DB entities with LIMIT operation. Such as

$query = 'SELECT * FROM HeroStats ORDER BY Heroname LIMIT 5'

You can use mysqli to connect Database

$db_connection = new mysqli([HOST], [USERNAME], [PASSWORD], [DATABASE_NAME]);

Use mysqli::fetch_array to fetch images to an array. This function will help you

function fetch($query)
{
    $res = $db_connection->query($query);
    $array = [];
    while( $r = $res->fetch_array(MYSQLI_ASSOC) )
    {
        array_push($array, $r);
    }

    return $array;
}

To pass array into Javascript, encode array to JSON or xml
```
$array = fetch($query);
$json = json_encode($array);
```
So, you can receive this response in Javascript, just convert JSON to array
```
var images = JSON.parse(response);
```

answered Dec 30, 2014 at 21:10

Vitalii

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2 Comments

Bogdan Molinger Over a year ago

can i connect to an MS Access database using mysqli ?

Vitalii Over a year ago

@BogdanMolinger, no, you can't, but you're didn't say it. If you have error Undefined index this means that array don't have element with such index. To debug Db result, please var_dump($heroes);die; after connection close

score 0 · Accepted Answer · 2014-12-30 21:34:59Z

0

I can see two errors in your code.

First:

Change this

$herolist[] .= $heroimage . ".png";

to

$herolist[] = $heroimage . ".png";

Second:

Remove .png from img src attribute and update for loop to use array length

for ($i = 0; $i < 5; $i++) {
  echo " <img src='Hero icons/$herolist[$i].png' class='heroimage'>";
}

to

for ($i = 0; $i < count($herolist); $i++) {
   echo " <img src='Hero icons/$herolist[$i]' class='heroimage'>";
}

Also, you should not use spaces for folder names. Change your folder name from Hero icons to Hero_icons and update the img tag.

edited Dec 30, 2014 at 21:34

answered Dec 30, 2014 at 21:22

user2616876

7 Comments

Bogdan Molinger Over a year ago

i've made the changes , now it prints out the correct path to the first hero from the team but not for the other ones , for example if first hero is a it prints <img src="Hero icons/a.png' class='heroimage'> but for the other ones only <img src="Hero icons/' class='heroimage'> and it says that it's an Undefined offset

user2616876 Over a year ago

Your query probably fetches only one image. Update the for loop as above.

Bogdan Molinger Over a year ago

The Undefined offset notices have disappeared but the images are still not printed , and it still prints out the code for the first image only

user2616876 Over a year ago

The query fetches only one image. Check your database and make sure it has all the images.

user2616876 Over a year ago

I'm not much familiar with Access database, but you can check the difference between the entry that shows and other entries and figure out why the conditions in your code don't satisfy.

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