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<?php
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//$file_dir="/htdocs/upload";
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// Connect to database
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$db_name1="test";// Database name
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$conn1=mysql_connect("localhost","","") or die("I Couldn't connect");
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$db=mysql_select_db($db_name1,$conn1) or die("I Couldn't select your database");
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data base
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//$q = <<<CREATE
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//create table pix (
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//pid int primary key not null auto_increment,
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//title text,
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// imgdata longblob)
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//CREATE;
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//(@mysql_query($q);
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-
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$imgdir="../image/latest.jpg";
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if ($_REQUEST['completed'] == 1) {
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// Need to add - check for large upload. Otherwise the code
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// will just duplicate old file ;-)
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// ALSO - note that latest.img must be public write and in a
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// live appliaction should be in another (safe!) directory.
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move_uploaded_file($_FILES['imagefile']['tmp_name'],$imgdir);
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$instr = fopen($imgdir,"rb");
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$image = addslashes(fread($instr,filesize($imgdir)));
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//echo "jamal khan =".$fupload;
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if (strlen($instr) < 149000) {
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mysql_query ("insert into pix (title, imgdata) values (\"".
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$_REQUEST['whatsit'].
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"\", \"".
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$image.
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"\")");
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} else {
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$errmsg = "Too large!";
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}
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}
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-
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$gotten = @mysql_query("select * from pix order by pid desc limit 1");
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if ($row = @mysql_fetch_assoc($gotten)) {
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$title = htmlspecialchars($row['title']);
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$bytes = $row['imgdata'];
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//echo"if executed succefully";
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} else {
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$errmsg = "There is no image in the database yet";
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$title = "no database image available";
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// Put up a picture of our training centre
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$instr = fopen($imgdir,"rb");
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$bytes = fread($instr,filesize($imgdir));
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}
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-
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if ($_REQUEST["gim"] ==1) {
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header("Content-type: image/jpeg");
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echo "sdfds";
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print $bytes;
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exit ();
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}
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?>
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<html><head>
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<title>Upload an image to a database</title>
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<body bgcolor=white><h2>Here`s the latest picture</h2>
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<font color=red><?= $errmsg ?></font>
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<center><img src=?gim=1 width=144><br>
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<b><?= $title ?></center>
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<hr>
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<h2>Please upload a new picture and title</h2>
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<form enctype=multipart/form-data method=post>
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<input type=hidden name=MAX_FILE_SIZE value=150000>
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<input type=hidden name=completed value=1>
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Please choose an image to upload: <input type=file name=imagefile><br>
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Please enter the title of that picture: <input name=whatsit><br>
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then: <input type=submit></form><br>
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<hr>
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</body>
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</html>
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This my source code it insert picture into data base but cannot read data from data base.
The Question is now that how i can i access Image using the above scnario
The Actual Problem here <body bgcolor=white>< h2>Here's the latest picture</h2>
<font color=red><?= $errmsg ?></font>
<center><img src=?gim=1 width=144><br>
<b><?= $title ?></center>
<hr>
in <img >tage but cannot understand.
Please help me, i will be very thankfull to all of u?
1  3523  Atli 5,058
Recognized Expert Expert
Hi.
What are you trying to do with the source parameter of the <img> tag? -
<img src=?gim=1 width=144>
-
The src parameter should contain an URL to an image, or a CGI script that outputs the contents of an Image.
The way you use it is as you want it to load the same page it is on, adding the 'gim=1' GET value.
If that is the case, and you want this code to both output the image data, and output the HTML form to add a new image, I would recommend against it.
It would be far easier to create two .php files. One of them to print and process the form data, as well as fetch the image title and such info, and the second .php file to fetch and print the image data. Then you could reference the second .php file from the first file.
P.S.
Please use [code] tags when posting code. It is impossible to read without them.
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