Error: <?php echo variable in string

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JanMolendijk
Posts: 107
Location: Holland Rotterdam

Error: <?php echo variable in string

I`m buissy with an simple upload script, every member have a own folder for pictures on my server.
Now i try to add a echo variable but getting a error:

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Warning: move_uploaded_file(/images/<?php echo $users[email];?>/689886.jpg): failed to open stream: No such file or directory in C:\xampp\htdocs\Comments\addnew.php on line 67
I have a part of my script in code

Code: Select all

<?php $url = 'http://'.$_SERVER['HTTP_HOST'].$_SERVER['PHP_SELF']; ?> <?php include('connection.php'); $id = (int) $_GET['id']; $sql=mysqli_query($conn,"SELECT * FROM `user` WHERE id='" . $id . "' "); $users=mysqli_fetch_assoc($sql); ?> <?php echo $users['email'];?> <?php error_reporting( ~E_NOTICE ); // avoid notice require_once 'dbconfig.php'; if(isset($_POST['btnsave'])) { $username = $_POST['user_name'];// user name $userjob = $_POST['user_job'];// user email $imgFile = $_FILES['user_image']['name']; $tmp_dir = $_FILES['user_image']['tmp_name']; $imgSize = $_FILES['user_image']['size']; if(empty($username)){ $errMSG = "Please Enter Username."; } else if(empty($userjob)){ $errMSG = "Please Enter Your Job Work."; } else if(empty($imgFile)){ $errMSG = "Please Select Image File."; } else { $upload_dir = '/images/<?php echo $users[email];?>/'; // upload directory $imgExt = strtolower(pathinfo($imgFile,PATHINFO_EXTENSION)); // get image extension
is their an different variable for: <?php echo $users[email];?>
- or how can i make it work into $upload_dir ?

Admin
Hello,
Not add <?php echo ; in the value of a variable; just concatenate with dot '.':

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$upload_dir ='/images/'. $users['email'] .'/';

JanMolendijk
Thanks it is working now if you wanna have a little donation how can i ?

Admin
Thanks for your intention, but I not need donation.
I'm thankful with my peace.

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