display image in my page using echo

Question

this is code of the page that I will get the image from(work perfectly)

<?php
    ob_start();
    session_start();
    include('connect.php');

    $id = $_GET['id'];
    $query = mysql_query("SELECT * FROM news WHERE id=$id");
    $row = mysql_fetch_assoc($query);

    header("Content-type: image/jpeg");
    echo $row['image'];
?>

and this is my page that i get the image to in

<?php
    ob_start();
    session_start();
    include('includes/connect.php');
    include('includes/phpCodes.php');

    $id = $_GET['id'];

    function showNews() {
        $data = array( 'id' => $id );
        $base = "includes/getImage.php";
        $url = $base. "?" . "id=36";
        echo $url;
        echo '<img src=includes/getImage.php class="newsImage">';
        echo '<h1><p class="subjecTitle">هنا العنوان</p></h1>   
                 <div class="newsContent">
                    hihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihi
             </div>
        ';
    }
?>

<!DOCTYPE html>
<html>
    <head>
        <title>عينٌ على الحقيقة</title>

        <meta charset="utf-8">

        <link rel="stylesheet" type="text/css" href="css/mainstyle.css">
        <link rel="stylesheet" type="text/css" href="css/showstyle.css">
        <script lang="javascript">
            function logout( myFrame ) {
                myFram.submit();
            }
        </script>
    </head>
    <body>
        <div class="wrapper">
            <?php headerCode(); ?>
            <div class="content" dir="rtl">
                <?php showNews(); ?>
            </div>
        </div>
    </body>
</html>

i think my wrong is in , can someone tell me how can I solve it?, sorry for my bad english

@BassamBadr now see http://stackoverflow.com/questions/17940111/display-image-in-my-page-using-echo/17940260#17940260 — KarSho, Jul 30 '13 at 07:18

Remko · Accepted Answer · 2013-07-30T07:49:11.550

1

Cleared it up for you:

echo '<img src="includes/getImage.php?id=' . $id . '" class="newsImage">';

Should 100% work (if the $id param has a value of course).

Update to fix the missing $id var:

    <?php
        ob_start();
        session_start();
        include('includes/connect.php');
        include('includes/phpCodes.php');

        $id = $_GET['id'];

        function showNews(){
            $id = $_GET['id'];
            $base = "includes/getImage.php";
            $url = $base. "?" . "id=36";
            echo $url;
            echo '<img src="includes/getImage.php?id=' . $id . '" class="newsImage">';

            echo '
                <h1><p class="subjecTitle">??? ???????</p></h1> 
                <div class="newsContent"></div>
            ';
        }
    ?>

edited Jul 30 '13 at 07:49

answered Jul 30 '13 at 07:13

Remko

968
6
19

Parse error: syntax error, unexpected '>' in C:\AppServ\www\Eye\show.php on line 15 – Bassam Badr Jul 30 '13 at 07:19
Are you positive your source looks exactly the same as provided by me? This will work and will not generate an error. Please show me your code as it looks like now. – Remko Jul 30 '13 at 07:23
See this working example: http://phpfiddle.org/lite/code/2h3-7ck and press 'run' to the right. Will result in a img tag without parsing errors. Are you sure you have exactly the same string on line 15? – Remko Jul 30 '13 at 07:28
it works but the image didn't display – Bassam Badr Jul 30 '13 at 07:43
the sourc look like this – Bassam Badr Jul 30 '13 at 07:45
The $id var was empty in that case and I see why (see my updated code) – Remko Jul 30 '13 at 07:46
I would like to add that the code is not safe for exploits (using unescaped $_GET data) see also: http://stackoverflow.com/questions/1695973/php5-security-get-post-parameters – Remko Jul 30 '13 at 07:51
after 10 hours of work :) Now it's work im so so so happy, thank you so much brother – Bassam Badr Jul 30 '13 at 08:00

Prasanth Bendra · Answer 2 · 2013-07-30T07:32:44.247

0

Change :

echo ' <img src=includes/getImage.php class="newsImage">';

To :

echo ' <img src="includes/getImage.php?id='.$id.'" class="newsImage">';

Please note src has " and also make sure that includes/getImage.php return a image path

edited Jul 30 '13 at 07:32

answered Jul 30 '13 at 06:54

Prasanth Bendra

31,145
9
53
73

I want to sent id as parameter to getImage.php – Bassam Badr Jul 30 '13 at 06:55
Just send it like this : `includes/getImage.php?id=1` in `getImage.php` use `$_GET['id']` to get this id – Prasanth Bendra Jul 30 '13 at 06:56
convert it to ; still get nothing – Bassam Badr Jul 30 '13 at 07:03
check in firebug or view souce and see what is the value for `src` in source – Prasanth Bendra Jul 30 '13 at 07:05
the source is ; – Bassam Badr Jul 30 '13 at 07:08
@BassamBadr try my code. – KarSho Jul 30 '13 at 07:10
If you execute the `includes/getImage.php?id=1` from browser are you getting the path of the image ? – Prasanth Bendra Jul 30 '13 at 07:11
@Prasanth Bendra im using google chrome when i open a show.php I pressed f12 and the copy the source – Bassam Badr Jul 30 '13 at 07:13
@BassamBadr see my comment because the $id param is not parsed by PHP in your current setup – Remko Jul 30 '13 at 07:14

score 0 · Answer 3 · answered Jul 30 '13 at 07:31

0

why are you using two different pages? put both code in single page and simply do this

 <img src=includes/<?php echo $row['image']; ?> class="newsImage">

answered Jul 30 '13 at 07:31

chirag ode

950
7
15

display image in my page using echo

3 Answers3