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in your form tag one attribute is missing. Without that attribute browser wont upload file
enctype="multipart/form-data"
I know I am coward since the day I know that fortune favors the brave
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johny10151981 wrote: in your form tag one attribute is missing. Without that attribute browser wont upload file
enctype="multipart/form-data"
Thanks for this.
I have got it working with uploadify [uploadify].
I am very grateful.
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I think this belongs better in the forum as I believe the problem to be on the PHP side. What I'm doing is creating a table dynamically that has some JavaScript function calls in it (well I'm trying to). The page is displaying ok in Chrome and Firefox but IE is complaining about it and not displaying at all. However in Chrome and Firefox it's not "working" just displaying right. this is the code I'm trying to use...
$display_string .= "<td><a href='javascript:choice('".$row .$col ."')'><img src='images/o.jpg' border=0 height=100 width=100 name='" .$row .$col ."' alt='Open'></a></td>";
When you hover over one of the cells on the loaded page you see javascript:choice(. It should read something like javascript:choice(A3) If I do an inspect element with chrome I get
<img src="images/o.jpg" border="0" height="100" width="100" name="B2" alt="Open"> so it appears the issue is in the creation of the javascript call?
Thanks in advance!!
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You need to properly match up your quotes - you used a single quote for href= and then the closing quote was just inside the javascript function. Try this:
$display_string .= "<td><a href=\"javascript:choice('{$row}{$col}')\"><img src='images/o.jpg' border=0 height=100 width=100 name='{$row}{$col}' alt='Open'></a></td>";
I've replaced the outer single quotes with escaped double quotes. I've also put the $row and $col variables into the string so that the '.' operator is not required, but that's just my personal preference.
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Now I see what I did wrong in my other version, I forgot the escape character... the joys of learning new languages. Thanks, it seems to be working now.
Putting a variable inside {} adds it to a string? I didn't know that. 
I really appreciate the help!!
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Thanks for the link, that really helps
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At the moment, I have index.php located at the root. I want to have it parse all my URLs and delegate the workload to the correct code (via URL rewriting). I may be a bit nit-picky, but I'd like to know if there's an elegant solution to neatly separating the request URL and the key-value pairs of the query?
My index.php looks like this:
<?php
header('Content-Type: text/plain; charset=utf-8');
echo 'Debug Output';
echo "\n\n";
echo 'Request ($_SERVER["REQUEST_URI"]): "'.$_SERVER['REQUEST_URI'].'"';
echo "\n";
echo 'Request ($_GET["request"]): "'.$_GET['request'].'"';
echo "\n";
echo 'd="'.$_GET['d'].'"';
echo "\n";
echo 'e="'.$_GET['e'].'"';
echo "\n";
?>
If I give it some URLs, they produce the corresponding outputs:
- mysite.com/a/b/c?d=1&e=2&request=3
Debug Output
Request ($_SERVER["REQUEST_URI"]): "/a/b/c?d=1&e=2&request=3"
Request ($_GET["request"]): "3"
d="1"
e="2" - mysite.com/a/b/c?d=3&e=4
Debug Output
Request ($_SERVER["REQUEST_URI"]): "/a/b/c?d=3&e=4"
Request ($_GET["request"]): "a/b/c"
d="3"
e="4"
$_SERVER["REQUEST_URI"] includes everything after mysite.com including the raw query string, and $_GET["request"] will not return the requested URL minus the query string if there is a key named request.
I could parse the value of $_SERVER["REQUEST_URI"] myself, but I'd like to use built-in functionality before coding around the problem.
My GUID: ca2262a7-0026-4830-a0b3-fe5d66c4eb1d 
Now I can Google this value and find all my Code Project posts!
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$_SERVER["QUERY_STRING"] will give you the query string. (I forget if it includes the ? or not.) Picking off everything prior to the first ? (i.e. the URL part) isn't too hard.
Cheers,
Peter
Software rusts. Simon Stephenson, ca 1994.
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Interestingly enough, I get all the pieces of the request neatly delimited by ampersands.
mysite.com/a/b/c?d=1&e=2&request=3
Debug Output
Request ($_SERVER["REQUEST_URI"]): "/a/b/c?d=1&e=2&request=3"
Request ($_GET["request"]): "3"
Request ($_SERVER["QUERY_STRING"]): "request=index.php&request=a/b/c&d=1&e=2&request=3"
d="1"
e="2"
mysite.com/a/b/c?d=1&e=2&request=%26+%3F
Debug Output
Request ($_SERVER["REQUEST_URI"]): "/a/b/c?d=1&e=2&request=%26+%3F"
Request ($_GET["request"]): "& ?"
Request ($_SERVER["QUERY_STRING"]): "request=index.php&request=a/b/c&d=1&e=2&request=%26+%3F"
d="1"
e="2"
Although the key-value pairs remain encoded, I guess it's still better to just use explode to get the key-value pairs first and then explode the resulting items to get the keys and values.
My GUID: ca2262a7-0026-4830-a0b3-fe5d66c4eb1d
Now I can Google this value and find all my Code Project posts!
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I did some testing, and it appears that certain values in the query can replace $1 in .htaccess with just 406.shtml or 406_shtml. mysite.com/a/b/c?d=1&e=2&%85 will cause this problem. It looks like I will have to parse the contents of $_SERVER['REQUEST_URI'] within the script if I want to handle all the cases without fail.
My GUID: ca2262a7-0026-4830-a0b3-fe5d66c4eb1d
Now I can Google this value and find all my Code Project posts!
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I have to analyze a system written in php. I am a .NET programmer. I will be emulating the php system. I have been given a tar dump of the directory. After decompression I need to find the starting php script. What should I be looking for? I have many php scripts, but it feels like a haystack. Can someone tell me what the starting needle name should be??
Thanks in advance... Chuck..
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the default page in a directory normally is called index.php
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Luc, Nice article. The problem is not recognising the PHP code or being able to read the code.
The problem is that I have several hundred PHP files. Which one is the first? That starts the app running. From there I and untangle the ball of string I have for years. Where is the beginning?
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Do all PHP programs start with "main.php" or "index.php" or "default.php"?
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No. In a web application, the default file in a folder is index.php, that is what gets served to a client when accessed by a browser.
You can create any filename.php you like and execute it locally by using a command line that fits your PHP platform and refers to the source file.
And there are ways to compile a bunch of PHP files into a single EXE, haven't done that yet.
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i want to learn python language in my mobile.. Can u give me information about these..
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use pys60
gomobile package,
free development you can download from sourceforce..
enjoye
all the best
http://dhruvdave.com
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anybody has bplustree code in python
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Please explain more detailed on the blender part coding. what is expected.
"Take chances, make mistakes. That's how you grow. Pain nourishes your courage. You have to fail in order to practice being brave." - Mary Moore
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Hi there,
Please can someone help me with this issue:
How do I display an image from a database showing the latest added images first (basically showing the last index to the first)
I have the rest of the code what I need is how do I access the files ?
eg:
<?php
<a href="?"><img src="?" alt="" width="50px" height="50px"/></a>
?>
please help where I have put the '?'.
The database is called pagination, the table is called images. There are 3 columns in the table `images`, name_id, name, image.
I want to display it in descending order acording to name_id.
If you could even coment with a link to a tutorial that would help I would appretiate it.
Thanx alot,
Josh
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You will have to create a seperate PHP file that is able to load the image from the database and output it to the browser. You then set the '?' in the sample code you gave to that PHP file.
The PHP script that will actually present the image to the browser will be something similar to this:
<?php
mysql_connect("localhost","username","xxxxxxx");
mysql_select_db("database");
$rs = mysql_query("select image from images order by name_id ASC limit 1");
$row = mysql_fetch_assoc($rs);
$imagebytes = $row['image'];
header("Content-type: image/jpeg");
print $imagebytes;
?>
Note that you have to set the correct content type or the browser won't know what it is receiving. A full tutorial on storing images in a DB can be found at http://phpro.org/tutorials/Storing-Images-in-MySQL-with-PHP.html[^]
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