如何正确使用content-type：image / jpeg？

Question

I've created a page "student_picture.php" using the content-type:image/jpeg. im having problem when I want to add other text in that same page..

here is the sample of my code:

<?php
session_start();

if(!isset($_SESSION["StudentNo"])){
    header("location:login.php");
}

$StudentNo = $_SESSION['StudentNo'];

require("includes/connection.php");

$sql = "SELECT StudentPicture from dbo.Students where StudentNo = '$StudentNo'";
$stmt = sqlsrv_query( $conn, $sql );

if( $stmt === false) {
    die( print_r( sqlsrv_errors(), true) );
}

while( $row = sqlsrv_fetch_array( $stmt, SQLSRV_FETCH_ASSOC) ) {
    $img = $row['StudentPicture'];

    if ($img == null ) {
        echo "<img src='img/default_pic.gif'>";
    } else {
        $img =  trim($img);
        header('Content-Type: image/jpeg');
        echo $img;
    }

    echo $StudentNo;
}
?>

The image is successfully displaying but the echo $StudentNo is not displaying.. can anyone help me with my prob? thanks in advance.

Answer 1

You have to exit your script, because the header is sent to the client browser. If the $_SESSION["StudentNo"] is not present the script will process and will try to ouput your image.

<?php
    session_start();
    if(!isset($_SESSION["StudentNo"])){
        header("location:login.php");
        die();
    }

And that is not a proper way to output an image, echo "<img src='img/default_pic.gif'>"; with header('Content-Type: image/jpeg'); is not ok, either you return without that header or you read the image:

if ($img == null){
    $img = 'img/default_pic.gif';
} else {
    $img =  trim($img);
}

header('Content-Type: image/jpeg');
readfile($img);

or you lose the header tag

if ($img == null){
    $img = "<img src='img/default_pic.gif'>";
} else {
    $img = "<img src='".trim($img)."'>";
}
echo $img;

So your script could be:

<?php
    session_start();
    if(!isset($_SESSION["StudentNo"])){
        header("location:login.php");
        die();
    }
    $StudentNo = $_SESSION['StudentNo'];

    require("includes/connection.php");

    $sql = "SELECT StudentPicture from dbo.Students where StudentNo = '$StudentNo'";
    $stmt = sqlsrv_query( $conn, $sql );
    if( $stmt === false) {
        die( print_r( sqlsrv_errors(), true) );
    }

    while( $row = sqlsrv_fetch_array( $stmt, SQLSRV_FETCH_ASSOC) ) {
        $img = $row['StudentPicture'];
        if ($img == null){
            $img = "<img src='img/default_pic.gif'>";
        } else {
            $img = "<img src='".trim($img)."'>";
        }
        echo $img;
        echo $StudentNo;
    }
?>

I don't know the functions inside your DB connection, but I think you should use something like $row = sqlsrv_fetch_row( $stmt, SQLSRV_FETCH_ASSOC); and make without that while