Programming Forums > Web Development and Design > PHP

woozy · Mar 15th, 2007, 4:36 PM

Let's suppose I capture a bunch of raw data to a variable.

For example:

//page.php
   $photo = fread(fopen($_FILES[$str]['tmp_name'], "r"), $_FILES[$str]['size']);
   $photo_type = $_FILES[$str]['type'];

captures the binary data for a file submitted by a form.

or

(Toggle Plain Text)

//page.php
$query = "SELECT photo FROM messages WHERE id = $id";
$conn = connect();
$result = $conn->query($query);
while ($row = $result->fetchRow()){$image = $row[0];}

will collect data of 'photo' is a "BLOB" field in a $sql database.

Okay, so let's suppose I've got $photo (a huge long string of data) and I've got $photo_type (the mime type for whatever type of image it is).

Now I could display this as an image by doing:

(Toggle Plain Text)

//page.php
   $photo = fread(fopen($_FILES[$str]['tmp_name'], "r"), $_FILES[$str]['size']);
   $photo_type = $_FILES[$str]['type'];


header('Content-Type: $photo_type);
$photo;

That's all fine and good but I don't want page.php to be the image itself. I want it to have the image in it as an <img> tag like:

(Toggle Plain Text)

//page.php
   $photo = fread(fopen($_FILES[$str]['tmp_name'], "r"), $_FILES[$str]['size']);
   $photo_type = $_FILES[$str]['type'];

?>
Hello <?= $name ?><br>
<img src="<? some code to render image ?>

My question is, how can I pass the $photo (and $photo_type) $variables to whatever "?some code to render image ?" is.

I've made the following workaround:

(Toggle Plain Text)

//page.php
   $photo = fread(fopen($_FILES[$str]['tmp_name'], "r"), $_FILES[$str]['size']);
   $photo_type = $_FILES[$str]['type'];

  $_SESSION['binary'] = $photo;
  $_SESSION['type'] = $type;
?>
Hello <?= $name ?><br>
<img src="image.php">

//image.php
get_session();
header("Content-Type: $_SESSION['type']");
$_SESSION['binary'];

but I'd prefer to pass the data directly. (Can I send them with POST if it's not a form response?)

Do I actually have to pass the data? The reason for the HTML code <img src=url> is simply to make an http request and recieve the response. Is there any way I can use PHP to bypass the request entirely and give the response (which would be "Content-Type: $type\n\n$photo") directly?

Any help appreciated,
wooz

Mar 15th, 2007, 4:36 PM	#1
woozy Newbie Join Date: Mar 2007 Posts: 15 Rep Power: 0	passing binary (or long) data (dynamic images) Let's suppose I capture a bunch of raw data to a variable. For example: (Toggle Plain Text) //page.php $photo = fread(fopen($_FILES[$str]['tmp_name'], "r"), $_FILES[$str]['size']); $photo_type = $_FILES[$str]['type']; //page.php $photo = fread(fopen($_FILES[$str]['tmp_name'], "r"), $_FILES[$str]['size']); $photo_type = $_FILES[$str]['type']; captures the binary data for a file submitted by a form. or (Toggle Plain Text) //page.php $query = "SELECT photo FROM messages WHERE id = $id"; $conn = connect(); $result = $conn->query($query); while ($row = $result->fetchRow()){$image = $row[0];} //page.php $query = "SELECT photo FROM messages WHERE id = $id"; $conn = connect(); $result = $conn->query($query); while ($row = $result->fetchRow()){$image = $row[0];} will collect data of 'photo' is a "BLOB" field in a $sql database. Okay, so let's suppose I've got $photo (a huge long string of data) and I've got $photo_type (the mime type for whatever type of image it is). Now I could display this as an image by doing: (Toggle Plain Text) //page.php $photo = fread(fopen($_FILES[$str]['tmp_name'], "r"), $_FILES[$str]['size']); $photo_type = $_FILES[$str]['type']; header('Content-Type: $photo_type); $photo; //page.php $photo = fread(fopen($_FILES[$str]['tmp_name'], "r"), $_FILES[$str]['size']); $photo_type = $_FILES[$str]['type']; header('Content-Type: $photo_type); $photo; That's all fine and good but I don't want page.php to be the image itself. I want it to have the image in it as an <img> tag like: (Toggle Plain Text) //page.php $photo = fread(fopen($_FILES[$str]['tmp_name'], "r"), $_FILES[$str]['size']); $photo_type = $_FILES[$str]['type']; ?> Hello <?= $name ?><br> <img src="<? some code to render image ?> //page.php $photo = fread(fopen($_FILES[$str]['tmp_name'], "r"), $_FILES[$str]['size']); $photo_type = $_FILES[$str]['type']; ?> Hello <?= $name ?><br> <img src="<? some code to render image ?> My question is, how can I pass the $photo (and $photo_type) $variables to whatever "?some code to render image ?" is. I've made the following workaround: (Toggle Plain Text) //page.php $photo = fread(fopen($_FILES[$str]['tmp_name'], "r"), $_FILES[$str]['size']); $photo_type = $_FILES[$str]['type']; $_SESSION['binary'] = $photo; $_SESSION['type'] = $type; ?> Hello <?= $name ?><br> <img src="image.php"> //image.php get_session(); header("Content-Type: $_SESSION['type']"); $_SESSION['binary']; //page.php $photo = fread(fopen($_FILES[$str]['tmp_name'], "r"), $_FILES[$str]['size']); $photo_type = $_FILES[$str]['type']; $_SESSION['binary'] = $photo; $_SESSION['type'] = $type; ?> Hello <?= $name ?><br> <img src="image.php"> //image.php get_session(); header("Content-Type: $_SESSION['type']"); $_SESSION['binary']; but I'd prefer to pass the data directly. (Can I send them with POST if it's not a form response?) Do I actually have to pass the data? The reason for the HTML code <img src=url> is simply to make an http request and recieve the response. Is there any way I can use PHP to bypass the request entirely and give the response (which would be "Content-Type: $type\n\n$photo") directly? Any help appreciated, wooz

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Programming Forums > Web Development and Design > PHP

passing binary (or long) data (dynamic images)