How to echo

Question

I am getting error when I try to echo image in PHP. Here is my code:

$sql = "SELECT id, title, filename FROM images";
$result = $conn->query($sql);


if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
    $pathx = "directory/images/";

    echo '
    <img src = "'$pathx'.'$row["filename"]'">
    ';
    //filenames in database end with appropriate image extensions like .png

}

And I am getting this error: Parse error: syntax error, unexpected '$pathx' (T_VARIABLE), expecting ',' or ';' in C:\xampp\htdocs\display.php on line 29

I have checked solutions from similar questions from here on Stackoverflow and implemented some of the accepted answers still my problem is not solved.

meanwhile something like the code below works well. So why is the one above not working?

echo'<img src = "directory/images/girl-g2109_5_720.jpg">';

`$pathx = "directory/images/"; $file = $row["filename"]; echo '';` — Abdulla Nilam, Dec 07 '17 at 05:19
My question has been marked duplicate. The answer supposed to be the original is quite an advanced answer that will be helpful to beginners only if they are specifically told what is wrong with their code. The original answer made sense to me because I first got the answer from this post — Elizabeth Kof, Dec 07 '17 at 05:50

score 4 · Accepted Answer · answered Dec 07 '17 at 05:21

4

Do like this

$pathx = "directory/images/";
$file = $row["filename"];

echo '<img src="'.$pathx.$file.'">';

Output

<img src="directory/images/girl-g2109_5_720.jpg">

answered Dec 07 '17 at 05:21

Abdulla Nilam

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Happy to help :) – Abdulla Nilam Dec 07 '17 at 05:37

score 1 · Answer 2 · answered Dec 07 '17 at 05:22

if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
    $pathx = "directory/images/";
    echo '<img src = "'.$pathx.$row["filename"].'">';
}

OR use can also use following way to render img tag

if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
    $pathx = "directory/images/"; ?> 
    <img src = "<?php echo $pathx.$row["filename"] ?>">
<?php } ?>

score 1 · Answer 3 · answered Dec 07 '17 at 05:23

It looks like there is an issue with the array concatenation.

The string '<img src = "' and variable $pathx are not concatenated properly.

There should be a concatenation operator ('.') in between.

Change the code to the following format to fix the issue.

Code

echo '<img src = "' . $pathx . $row["filename"] . '">';

Output

<img src="directory/images/girl-g2109_5_720.jpg">

score 1 · Answer 4 · answered Dec 07 '17 at 05:35

Try this

There are multiple ways to display the image. You can also use like this.

if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
    $image=$row["filename"];
?>
<img src="directory/images/<?php echo $image;?>">
<?php
}
 ?>

Output

<img src="directory/images/image_name.png">

How to echo

4 Answers4

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