Innerhtml as php echo include

Question

I want to simply create function that on button click will change innerHTML of my div to one of the php files.

Example of one php file is like this:

<?php
    require('../Php/ConfigBaza.php');
    $sql = "
    SELECT Proizvod.PROIZVODID, Proizvod.NAZIV, Proizvod.SLIKA
    FROM PROIZVOD_GRUPE
    LEFT JOIN Proizvod ON PROIZVOD_GRUPE.PROIZVODID = Proizvod.PROIZVODID WHERE PROIZVOD_GRUPE.PROIZVOD_GRUPAID = '1';

    SELECT PROIZVODID, NAZIV, SLIKA FROM Proizvod WHERE AKCIJSKI_ARTIKAL = 1";

    $result = $mysqli->query($sql);

    if($result->num_rows < 1)
    {
        echo("Doslo je do greske prilikom ucitavanja proizvoda!");
        die();
    }
    else
    {

        echo("
        <div id='overlay' style='display: none; position: fixed; top: 0; bottom: 0; left: 0; right: 0; z-index: 999; background-color: hsla(0, 0%, 50%, 0.39);'>
            <div id='kolicina_overlay' style='display: none; position: fixed; width: 500px; height: 110px; margin-top: -55px; margin-left: -150px; top: 50%; left: 50%; text-align: center; background-color: #2196F3; z-index: 1000'>
                <p style='font-weight: bolder'>Unesite zeljenu kolicinu!</p>
                <input type='number' value='1' style='width: 50px'>
                <div id='kolicina_potvrdi_button' style='padding: 10px; color: white;' onclick='UnesiKolicinu()'>Potvrdi</div>
            </div>
        </div>
        ");
        while($proizvod = mysqli_fetch_assoc($result))
        {
            $pid = $proizvod['PROIZVODID'];
            $naziv = $proizvod['NAZIV'];
            $slika = $proizvod['SLIKA'];
            $rabat = $proizvod['POPUST'];

            if(!isset($slika) || empty($slika))
            {
                $slika = "https://upload.wikimedia.org/wikipedia/commons/thumb/a/ac/No_image_available.svg/600px-No_image_available.svg.png";
            }
            echo("
                <div class='proizvod_Box' id='$pid' onclick='IdiNaProizvod(id)'>
                    <div class='proizvod_Naslov'>
                        <p>$naziv</p>
                    </div>
                    <div class='proizvod_Slika'>
                        <img src='$slika'>
                    </div>
                </div>
            ");
        }
    }
?>

And how i tried to invoke them is like this:

case "_Akcija":
    document.getElementById('rightBox').innerHTML = "<?php echo include_once '../Php/Proizvodi/_Akcija.php';?>";
    break;

But it returns me error: Invalid or unexpected token for my ""

I have tried switching '' with "" inside my javascript but not working. Also tried to see maybe it will work with document.write but still same error. What to do?

you need to load the content with ajax first and then use innerHtml to print it — Elementary, Aug 14 '18 at 06:46
What’s causing the error is most likely that your PHP script outputs multi-line content, but “plain” JavaScript text literals can not go over multiple lines, that would need either ES6 syntax, or some manipulation of the content, see https://stackoverflow.com/questions/805107/creating-multiline-strings-in-javascript — CBroe, Aug 14 '18 at 07:11

score 0 · Answer 1 · answered Aug 14 '18 at 06:57

You should just include the php file on top of your client page, first change the echo to a variable:

<?php
        require('../Php/ConfigBaza.php');
        $sql = "
        SELECT Proizvod.PROIZVODID, Proizvod.NAZIV, Proizvod.SLIKA
        FROM PROIZVOD_GRUPE
        LEFT JOIN Proizvod ON PROIZVOD_GRUPE.PROIZVODID = Proizvod.PROIZVODID WHERE PROIZVOD_GRUPE.PROIZVOD_GRUPAID = '1';

        SELECT PROIZVODID, NAZIV, SLIKA FROM Proizvod WHERE AKCIJSKI_ARTIKAL = 1";

        $result = $mysqli->query($sql);

        if($result->num_rows < 1)
        {
            echo("Doslo je do greske prilikom ucitavanja proizvoda!");
            die();
        }
        else
        {

            echo("
            <div id='overlay' style='display: none; position: fixed; top: 0; bottom: 0; left: 0; right: 0; z-index: 999; background-color: hsla(0, 0%, 50%, 0.39);'>
                <div id='kolicina_overlay' style='display: none; position: fixed; width: 500px; height: 110px; margin-top: -55px; margin-left: -150px; top: 50%; left: 50%; text-align: center; background-color: #2196F3; z-index: 1000'>
                    <p style='font-weight: bolder'>Unesite zeljenu kolicinu!</p>
                    <input type='number' value='1' style='width: 50px'>
                    <div id='kolicina_potvrdi_button' style='padding: 10px; color: white;' onclick='UnesiKolicinu()'>Potvrdi</div>
                </div>
            </div>
            ");
            while($proizvod = mysqli_fetch_assoc($result))
            {
                $pid = $proizvod['PROIZVODID'];
                $naziv = $proizvod['NAZIV'];
                $slika = $proizvod['SLIKA'];
                $rabat = $proizvod['POPUST'];

                if(!isset($slika) || empty($slika))
                {
                    $slika = "https://upload.wikimedia.org/wikipedia/commons/thumb/a/ac/No_image_available.svg/600px-No_image_available.svg.png";
                }
                $myVar = "
                    <div class='proizvod_Box' id='$pid' onclick='IdiNaProizvod(id)'>
                        <div class='proizvod_Naslov'>
                            <p>$naziv</p>
                        </div>
                        <div class='proizvod_Slika'>
                            <img src='$slika'>
                        </div>
                    </div>
                ";
            }
        }
    ?>

Then include the php file on top of the page where you will use the javascript:

<?php echo include_once '../Php/Proizvodi/_Akcija.php';?>

<!-- Your javascript here below -->

After that, just echo the variable you defined on the file and the code should look like this:

<?php echo include_once '../Php/Proizvodi/_Akcija.php';?>
<script>
 switch(test){
  case "_Akcija":
    document.getElementById('rightBox').innerHTML = "<?php echo $myVar; ?>";
    break;
  default:
    break;
}
</script>

That should do the trick.

I do not want to include only var. I want whole file to include. My whole file is dipslaying creating and dsiplaying part of page — Pacijent, Aug 14 '18 at 07:00
You can include the whole content to the variable and just use the method on my answer, including too many php files is not really a good habbit and you should learn to organize your files properly. You can do it like this: $myVar = ""; $myVar .= "Content1"; $myVar .= "Content2"; Then you can use that variable to print all the contents you need. — Norielle Cruz, Aug 14 '18 at 07:02

Innerhtml as php echo include

1 Answers1