How do I solve this PHP error

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The error message:

syntax error, unexpected '$image_name' (T_VARIABLE)

The code:

if (isset($_POST['add-post'])) {
    adminOnly();
    $errors = validatePost($_POST);

    ($_FILES['image']['name'])
        $image_name = time() . '_' . $_FILES['image']['name'];

        $destination = ROOT_PATH . "/assets/images/" . $image1_name;

        $result = move_uploaded_file($_FILES['image1']['tmp_name'], $destination);

        if ($result) {
           $_POST['image1'] = $image1_name;

        } else {
            array_push($errors, "Failed to upload image");
        }
    } else {
       array_push($errors, "Post image required");
    }

What I have tried:

I have tried using other variables

Posted 10-Jul-20 1:24am

Brian Nyaberi

Updated 10-Jul-20 23:17pm

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2 solutions

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Xyzdesk · Answer 1 · 2020-07-10T01:36:00

You declared $image_name but used $image1_name
It also looks like you forgot your semicolon on line 7.

PHP

if (isset($_POST['add-post'])) {
    adminOnly();
    $errors = validatePost($_POST);

    ($_FILES['image']['name']);
        $image_name = time() . '_' . $_FILES['image']['name'];

        $destination = ROOT_PATH . "/assets/images/" . $image_name;

        $result = move_uploaded_file($_FILES['image1']['tmp_name'], $destination);

        if ($result) {
           $_POST['image1'] = $image1_name;

        } else {
            array_push($errors, "Failed to upload image");
        }
    } else {
       array_push($errors, "Post image d");
    }

Andre Oosthuizen · Answer 2 · 2020-07-10T23:17:00

Solution 2

You have declared

$image_name

yet you use

$image1_name

notice the 1 (one).

Posted 10-Jul-20 23:17pm

Andre Oosthuizen