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hi to all.. how are you? -
help. which code should i use? -
i used this:
Code:<input name="image" type="image" src="asdf.php?imgid=<?Php echo $form_no; ?>" width="110" height="100"/>
Code:error_reporting (E_ALL ^ E_NOTICE); $id = $_GET['imgid']; $Link = mysqli_connect("localhost","root","***", "***");
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with this code the data has inserted.. but now im having a problem retrieving the photo.....Leave a comment:
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yeah.. in atli's code there were files in the DB but when i tried to check my own, there was no files.. im going to start again.. hope i can do it.. i'll try to fix the codes.. the insert query maybe has an error..
here's now my code:
Code:<?php error_reporting (E_ALL ^ E_NOTICE); $pid = base64_decode($_GET['id']); if(isset($_FILES['uploaded_file'])) { if($_FILES['uploaded_file']['error']
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okei.. im trying a lot of codes that is why i am so confused.. the issue about the displaying picture from mysql is still unsolved..Leave a comment:
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yes
is this correct?
Code:$pic = $_FILES['Browse']; $formid = mysqli_query($Link, "Select form_no from registration_form where last_insert_id() = form_no"); $formnum = mysqli_fetch_row($formid); $querypic = mysqli_query($Link, "INSERT INTO pix(pid, imgdata) values('$formnum', '$pic')");
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i tried another code:
here's the code. it only displays a blank photo. i mean an x that is usually seen in a blank photo..
Code:<?php $Link = mysqli_connect("localhost","root","tupi", "amyak_maleh"); if (!$Link) { trigger_error("Could not connect", E_USER_ERROR ); } $result=mysqli_query($Link, "SELECT * FROM
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the image is not showing.. i tried to show it through the src itself and its okei but when the image that came from the database it is not showing,.. maybe the problem is in the database or in the code, but there is no sign error in the codes.Leave a comment:
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solution: put the form handling script in the action attribute.
what exactly i will put in the action attribute?Leave a comment:
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this is the code..
Code:<form action="image.php" method="post"> <input type="image" src="image.php?form_no=<?Php echo $form_no; ?>" /> </form>
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whats this?when i click the submit query which is the one appeared insted of the image, it jumps to the image.php page then this is what appears:
http://localhost/image.php?x=&y=Leave a comment:
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here's the code, and it appears only a this string: Submit Query
Code:<input type="image" src="image.php?fileid=<?Php $form_no; ?>" />
thanksLeave a comment:
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i see.. it should be in a different page? ..so that it will be a valid code in src?Leave a comment:
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how could i convert the strings into an image? what function should i use? to convert it?
thanksLeave a comment:
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i tried atli's manual and the picture is not appearing?
here's the code for retrieving the info in the pic..but the info in filedata only showing a set of strings..
Code:<?php $Link = mysqli_connect("localhost","root","tupi", "amyak_maleh"); if(mysqli_connect_errno()){ die("Mysql connection failed:".mysqli_connect_errno()); } //$result1 = mysqli_query($Link, "SELECT filedata FROM filestorage
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