for loop not working correctly with json encode Php

Question

I am trying to fetch members ratings from the database using ajax. I passed a function to the JSON, even though it returned a value in the function but it doesn't execute the for loop condition.

Here is my code, the loop failed to execute. What am I doing wrong?

    function mrate($irate) {
        $class = "fa-star star-filled";
        for ($i = 0; $i < 5; $i++) {
            if ($irate <= $i) {
                $class = "fa-star-o empty";
            }
            return '<i class="fa ' . $class . '"></i>';
        }
    }

    $perPage = 2;
    if (isset($_GET["page"]) && isset($_GET["page"])) {
        $page = $_GET["page"];
        $pid = $mysqli->real_string_escape($_GET["pid"]);
    } else {
        $page = 1;
        $pid = $mysqli->real_string_escape($_SESSION['pid']);
    };

    $startFrom = ($page - 1) * $perPage;
    $sqlQuery = "SELECT id, name,
        review, rating, added_date
        FROM review_rating
        where product_id = '$pid'
        ORDER BY id ASC LIMIT $startFrom, $perPage";
    $result = mysqli_query($mysqli, $sqlQuery);
    $paginationHtml = '';
    while ($row = mysqli_fetch_assoc($result)) {

        $img = '<img class="rounded-circle" width="50" src="' . $set['installUrl'] . 'assets/img/login.png" alt="' . $row["name"] . '"/>';
        $irate = $row['rating'];

        $paginationHtml .= '<div  class="product-review pb-4 mb-4 border-bottom">';
        $paginationHtml .= '<div class="d-flex mb-3">';
        $paginationHtml .= '<div class="media media-ie-fix align-items-center mr-4 pr-2">' . $img;
        $paginationHtml .= '<div class="media-body pl-3"><h6 class="font-size-sm mb-0">' . $row["name"] . '</h6>';
        $paginationHtml .= '<span class="font-size-ms text-muted">' . $row['added_date'] . '</span></div></div>';
        $paginationHtml .= '<div><div class="star-rating">' . mrate($irate) . '</div></div>';
        $paginationHtml .= '</div>';
        $paginationHtml .= '<p class="font-size-md mb-2">' . $row['review'] . '</p>';
        $paginationHtml .= '</div>';
    }
    $jsonData = array(
        "html" => $paginationHtml,
    );
    echo json_encode($jsonData);

Your for loop won't iterate more than once because of the return statement. — Rajdeep Paul
– Rajdeep Paul, Commented Mar 10, 2020 at 15:34
@RajdeepPaul, how could I return the value in the function without using retun? — Fagbemi Ayodele
– Fagbemi Ayodele, Commented Mar 10, 2020 at 15:39
Can you please explain what you are trying to do in mrate function, especially in that for loop? — Rajdeep Paul
– Rajdeep Paul, Commented Mar 10, 2020 at 15:42
@RajdeepPaul, I am trying to print out the number of rating per members. The star-icon appears 5 times. Out of the 5 stars, get number of rating per member which brings out the looping. — Fagbemi Ayodele
– Fagbemi Ayodele, Commented Mar 10, 2020 at 15:46

Umer Abbas · Accepted Answer · 2020-03-10 16:35:10Z

2

Replace your function mrate($irate) with this and try. You needed to concatenate the stars code to display it more than once.

function mrate($irate){
    $stars = '';
    for($i=0; $i<5; $i++){
        if($irate <= $i){
            $class = "fa-star-o empty";
        }else{
            $class = "fa-star star-filled";
        }
        $stars .= '<i class="fa '.$class.'"></i>';
    }
    return $stars;
}

edited Mar 10, 2020 at 16:35

answered Mar 10, 2020 at 15:55

Umer Abbas

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3 Comments

Nigel Ren Over a year ago

You could factor out the $stars .= ... part so that you don't repeat it in both branches of the if..else...

Umer Abbas Over a year ago

@NigelRen Thank you for the reminder, edited my answer.

Fagbemi Ayodele Over a year ago

Thanks to everyone who contributed. I appreciate you.

Rajdeep Paul · Accepted Answer · 2020-03-10 16:19:44Z

0

Assuming there's no fractional rating, you can do the following - Display all 5 stars but solid ones will represent the rating.

function mrate($irate){
    $class = '';
    for($i = 0; $i < 5; $i++){
        if ($irate <= $i) {
            $class .= '<i class="fa fa-star"></i>';
        } else {
            $class .= '<i class="fa fa-star-o"></i>';
        }
    }
    return $class;
}

answered Mar 10, 2020 at 16:19

Rajdeep Paul

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