Hi a bit of PHP help needed.

rofl90 · August 24, 2007

OK, so I've coded up this login and I've got a separate section and now I'm designing a website for an artist, and to my knowledge I don't know if It's possible to code a section where my client can browse for a picture and then fill out another form to write a comment, in the backend, and then to display the picture in a specified area in the frontend, could anyone help me out?

trq · August 24, 2007

What you'll want to do is store the image path and its related comment in a database. You then simply supply a form for your client to fill in there comment based on the record id.

If this goes over your head, you'll probably need to look at some tutorials as this sort of thing is well covered. You might try the link in my signiture.

rofl90 · August 24, 2007

I understand it so I just store the image path in MySQL then I pluck the image path into where it should be to display it in PHP, in PHP, and the comment where it should, be..

OK, I understand, also what's wrong with this syntax error line 23, I don't understand it

<?php

/**
* @Charles Kirk
* @copyright 2007
*/

ob_start();
include("config.php");
if($logged[username] && $logged[level] ==5)
{

if($_GET[user])
{

if (!$_POST[update])
{

$user = mysql_query("SELECT * from users where username = '$_GET[user]'");
$user = mysql_fetch_array($user);
echo("

<div align="center"><form method="POST">
<table width="100%">
<tr>
<td align="right" width="25%">
User Level 
</td>
<td align="left">
<input type="text" size="25" maxlength="25" name="level"
value="$user[level]"></td>
</tr>
<tr>
<td align="right" width="25%">
Location 
</td>
<td align="left">
<input type="text" size="25" maxlength="25" name="locate"
value="$user[location]"></td>
</tr>
<tr>
<td align="right" width="25%">
MSN Messenger 
</td>
<td align="left">
<input size="25" name="msn" value="$user[msn]"></td>
</tr>
<tr>
<td align="right" width="25%">
AOL Messenger</td>
<td align="left">
<input size="25" name="aim"  value="$user[aim]"></td>
</tr>
<tr>
<td align="right" width="25%">
Email Address</td>
<td align="left">
<input size="25"  name="email" value="$user[email]"></td>
</tr>
<tr>
<td align="center">
</td>
<td align="left">
<input type="submit" name="update" value="Update"></td>
</tr>
</table>
</form>
</div>");

}
else
{
$email = htmlspecialchars($_POST[email]);
$aim = htmlspecialchars($_POST[aim]);
$msn = htmlspecialchars($_POST[msn]);
$locate = htmlspecialchars($_POST[locate]);
$level = htmlspecialchars($_POST[level]);

echo ("$_GET[user]'s profile has been updated.");
$update = mysql_query("Update users set email = '$email',
msn = '$msn', aim = '$aim',
location = '$locate', level = '$level' where username = '$_GET[user]'");

}
}
else
{
$getusers = mysql_query("Select * from users order by username asc");
while($users = mysql_fetch_array($getusers))
{

echo("<a href="admin.php?user=$users[username]">$users[username]</a><br />");

}
}
}
else
{

echo("Sorry, but you are not allowed to view this page!");
}
?>

phpSensei · August 24, 2007

Post Line 23, and the error message.

rofl90 · August 24, 2007

Parse error: syntax error, unexpected T_STRING in D:\Web Projects\******\admin.php

teng84 · August 24, 2007

escape the " with \

look you use this

echo("<div align="center"><form method="POST">

should be

echo("<div align=\"center\"><form method=\"POST\">

or

echo ' <div align="center"><form method="POST">

note i dont include your whole script its just the logic i gave you

rofl90 · August 24, 2007

Still giving the same syntax error for both I'll give you 5 up and 5 down:

if (!$_POST[update])
{

$user = mysql_query("SELECT * from users where username = '$_GET[user]'");
$user = mysql_fetch_array($user);
echo("

echo    '   <div align="center"><form method="POST">
<table width="100%">
<tr>
<td align="right" width="25%">

or so...

teng84 · August 24, 2007

<?php

/**
* @Charles Kirk
* @copyright 2007
*/

ob_start();
include("config.php");
if($logged[username] && $logged[level] ==5)
{

if($_GET[user])
{

if (!$_POST[update])
{

$user = mysql_query("SELECT * from users where username = '{$_GET['user']}'");
$user = mysql_fetch_array($user);
echo'

<div align="center"><form method="POST">
<table width="100%">
<tr>
<td align="right" width="25%">
User Level 
</td>
<td align="left">
<input type="text" size="25" maxlength="25" name="level"
value="$user[level]"></td>
</tr>
<tr>
<td align="right" width="25%">
Location 
</td>
<td align="left">
<input type="text" size="25" maxlength="25" name="locate"
value="$user[location]"></td>
</tr>
<tr>
<td align="right" width="25%">
MSN Messenger 
</td>
<td align="left">
<input size="25" name="msn" value="$user[msn]"></td>
</tr>
<tr>
<td align="right" width="25%">
AOL Messenger</td>
<td align="left">
<input size="25" name="aim"  value="$user[aim]"></td>
</tr>
<tr>
<td align="right" width="25%">
Email Address</td>
<td align="left">
<input size="25"  name="email" value="$user[email]"></td>
</tr>
<tr>
<td align="center">
</td>
<td align="left">
<input type="submit" name="update" value="Update"></td>
</tr>
</table>
</form>
</div>';

}
else
{
$email = htmlspecialchars($_POST[email]);
$aim = htmlspecialchars($_POST[aim]);
$msn = htmlspecialchars($_POST[msn]);
$locate = htmlspecialchars($_POST[locate]);
$level = htmlspecialchars($_POST[level]);

echo ("{$_GET[user]}'s profile has been updated.");
$update = mysql_query("Update users set email = '$email',
msn = '$msn', aim = '$aim',
location = '$locate', level = '$level' where username = '{$_GET[user]}'");

}
}
else
{
$getusers = mysql_query("Select * from users order by username asc");
while($users = mysql_fetch_array($getusers))
{

echo("<a href="admin.php?user=$users[username]">".$users[username]."</a><br />");

}
}
}
else
{

echo("Sorry, but you are not allowed to view this page!");
}
?>

rofl90 · August 24, 2007

Works now same error for line 92

teng84 · August 24, 2007

use this

echo'<a href="admin.php?user="'.$users[username].'">'.$users[username].'</a><br />"';

rofl90 · August 24, 2007

Thanks

rofl90 · August 24, 2007

OK I've done the comments but how on earth do I start on uploading a picture file into mysql

phpSensei · August 24, 2007

edit: nvm,

rofl90 · August 24, 2007

could you show me what the code would look like

rofl90 · August 24, 2007

in plain jane i need to upload tiff files names and then i need to be able to show the file name in a js tag, sorry im new to php&mysql

phpSensei · August 24, 2007

crap, I just posted the code for you but said nvm, because i didnt know what you ment.

rofl90 · August 24, 2007

I mean couldyou post the code to upload images names to mysql, and to show them please?

rofl90 · August 24, 2007

.. I don't need the code to show it, just to upload it into table (images), btw I don't know if theres anything that goes on with image types, but it's Tiff

rofl90 · August 24, 2007

Sorry, I also have like 7 categories that my client must be able to sort his pictures into.

rofl90 · August 24, 2007

This is what I have so far..

<?php

/**
* @Charles Kirk
* @copyright 2007
*/

if ($_REQUEST[completed] == 1) {
       $source = "images";
       move_uploaded_file($_FILES['filename']['tmp_name'],
               "../$source/".$_FILES['filename']['name']);
       if (! eregi ('(gif|jpg|tif|pdf)$',$_FILES['filename']['name'])) {
       $fi = file("../$source/".$_FILES['filename']['name']);
       $fi2 = fopen("../$source/".$_FILES['filename']['name'],"w");
       foreach ($fi as $lne) {
               $n = rtrim ($lne);
               fputs ($fi2,"$n\n");
               }
       }
//
?>
Your file has been uploaded.
<?php } else { ?>
<br><br>
<form enctype=multipart/form-data method=post>
<input type=hidden name=MAX_FILE_SIZE value=150000>
<input type=hidden name=completed value=1>
Choose file to send: <input type=file name=filename> and
<input type=submit name=Submit></form><br>
<?php  } ?>
</body>
</html>

rofl90 · August 24, 2007

Anybody help me out?

phpSensei · August 24, 2007

So You Have 7 categories, and you want the author to be able to upload the picture into each 7?

rofl90 · August 24, 2007

I want the client to be able to upload the picture which i've done, then at the same time upload the name into mysql to take out and show in a php file, he also needs to be able to choose out of 7 categories, and he needs to be able to create more categories... greedy needy man

EDIT the file i posted earlier doesn't work eg it doesn't like add the picture to http://www.******.com/images/

rofl90 · August 24, 2007

Also ******** = website name

phpSensei · August 24, 2007

Okay, you have created the upload script, then:

Put in Mysql



$filename=$_FILE['filefield']['name'];

$data="INSERT INTO table (filename) VALUES ('$filename')";
$query=mysql_query($data);

Show it in a php file:



$data="SELECT * FROM table";
$query=mysql_query($data);

$row=mysql_fetch_assoc($query);

echo "<img src='uploaddirectory/" .$row['filefield']. "'>";

Thats All I understood from you..

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Hi a bit of PHP help needed.

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