Forms and Php and SQL

[B1011011]

Hello.

Let me explain what im doing since I dont know why the error is popping up...

I want to rate a picture and using a submit button assert the data into my database. I want to do this without having it refresh the page. The submit button seems to give me the errors.




I have an images file that displays a file based on what a user selects from the main page. Like this....

$output = "<img width=\"400\" src=\"images1/";
$output .= $file_name;
$output .= "\">";
echo $output;

After the image is displayed I have a rate the pictures form. Like this.


<form method="post" action="<?=$PHP_SELF?>">

<select>
<OPTION SELECTED VALUE="http://www.feistypics.com">Rate This Picture</OPTION>
<OPTION VALUE="0">1 - Bad </OPTION>
<OPTION VALUE="1">2 - Poor </OPTION>
<OPTION VALUE="2">3 - Ok </OPTION>
<OPTION VALUE="3">4 - Good </OPTION>
<OPTION VALUE="4">5 - Excellent </OPTION>
</select>

<input type="submit" name="submitrate" value="Submit">
</form>

<?
if ( $submitrate == Submit )
{
$count = mysql_query ("SELECT * FROM lookCount WHERE IDref = $id ");
$newcount = (($count + 3) /5);
mysql_query ("UPDATE lookCount set count = $newcount where IDref = $id ");
?>

Now Im not entirely sure what the "<?=$PHP_SELF?>"> line does so I need clarification on that but. When someone clicks submit... I get 2 different errors in my code.

http://www.feistypics.com Is the Main Page
http://www.feistypics.com/timages.php?id=133
Replace the "t" with "c" to see what Im working on i.e.
http://www.feistypics.com/cimages.php?id=133

If you try to submit using the cimages page it refreshes all wrong with errors. Im sure its simple solution but I havent found the answer.

Pls Assist.

[=X¥®µ§=]

first of all I'd like to point out that you can write this a bit shorter

$output = "<img width=\"400\" src=\"images1/";
$output .= $file_name;
$output .= "\">";
echo $output; 

to:

print "<img width=\"400\" src=\"images1/$file_name\">"; 

Now Im not entirely sure what the "<?=$PHP_SELF?>"> line does

by pressing the submit button, the action="" will tell the server what page to submit this data to.

so you could use "cimages.php" instead -> $PHP_SELF will give you the name of your current page

I didn't see any errors after submitting, didn't receive anything like "your vote has been submitted" either, so I'm not sure if it worked or not :/

[B1011011]

Thanks for clarifying that stuff up.

I know that the editings of the counter data isnt work.

Also after you choose submit, didnt you see it load http://www.feistypics.com/cimages.php

That page wont load without an ID being supplied to it.

When i click submit it goes to this cimages.php page. I dont want it to leave this page or I want to be able to send it to the same page it is on.

[FireFly]

1. Your <select> box doesn't have a name... it needs to be more like <select name="rate"> and then you will be able to use $rate in your script.

2. You need to somehow pass the ID of the picture in the form, like this - <input type="hidden" name="id" value="XXX"> - you change XXX with whatever is right according to your script.

3. This code:

$count = mysql_query ("SELECT * FROM lookCount WHERE IDref = $id "); 
$newcount = (($count + 3) /5); 

You need to:
a) mysql_fetch_array() the $count before you could use the information from that query.
b) Use $count['count'] in your formula, not just $count.

4. And last but not least, you need to quote Submit in the following line:

if ( $submitrate == Submit ) 

besides, I would use this instead (I don't know why, I just feel better with it ):

if ( $submitrate != '' ) 

Edit:
One more thing... in the formula for calculating the new count:

$newcount = (($count + 3) /5); 

shouldn't you use the value from the <select> box as well here?

[B1011011]

Excellent points. oversites on my part definitely. Nice Job.

Only part I am still sketchy about is how to pass the ID in the URL once I submit.

[=X¥®µ§=]

2 ways, either you do this by

<form method="post" action="<?=$PHP_SELF?id=$id?>">

or as firefly said, pass it on with a hidden field

2. You need to somehow pass the ID of the picture in the form, like this - <input type="hidden" name="id" value="XXX"> - you change XXX with whatever is right according to your script.

so <input type="hidden" name="id" value="$id">

this will result in the same image being showed...

[B1011011]

Made that change on the action="<?=$PHP_SELF?id=$id?>">

now i get parse error.

[=X¥®µ§=]

I'm not on my own box now so I can't test what would work, but there are a few possibilities
either

make
$url = $PHP_SELF . "?id=" . $id;
or
$url = "$PHP_SELF?id=$id";
or
$url = $PHP_SELF . "?id=$id";

and then do action="<?= $url ?>">

can't say by heart what would work, let me know

[B1011011]

Well those three all work for sending the correct URL but for some reason mySQL is fighting me.

Bunch of parse errors.

Weirdo... Gonna work on it a bit. Launch me an Email if you have an inspiration. gatorben@ufl.edu

[B1011011]

line thirty where i get the error has this code

$mysql_link = mysql_connect($dbhost,$dbuser,$dbpass);
$column = mysql_list_fields($dbname,$dbtable,$mysql_link);
$sql = "SELECT * FROM $dbtable where ID = $id";
$result = mysql_db_query($dbname,$sql);
while($value = mysql_fetch_array($result))
{
$desc = $value["description"];
$file_name = $value["file_name"];

[FireFly]

Originally posted by B1011011
Made that change on the action="<?=$PHP_SELF?id=$id?>">

now i get parse error.

I think that the <?=$foo ?> syntax only works for variables.

B1011011, what error exactly are you getting now? (and it would help to know what line 30 is )

[Defender1]

Originally posted by B1011011
Made that change on the action="<?=$PHP_SELF?id=$id?>">

now i get parse error.

If your not using PHP4 then that sytax for echoing a variable won't work.
try that line as <form method="post" action="<?php echo $PHP_SELF?id=$id; ?>"> and see if it works.

[FireFly]

<form method="post" action="<?php echo $PHP_SELF.'?id='.$id; ?>">

since "?id" is a string.

[Defender1]

And so isn't $PHP_SELF.
You only need to put quotes around string variables when running queries, not to echo values.
If you did, this wouldn't work

$var = "Hi!";
echo $hi; 

Though i didn't for the quotes for the echo statement.

<form method="post" action="<?php echo "$PHP_SELF?id=$id"; ?>">

[FireFly]

But you didn't put quotes around your echo statement:

Originally posted by Defender1
<form method="post" action="<?php echo $PHP_SELF?id=$id; ?>">

[B1011011]

That new line the post method doesnt work either.

errors are

Supplied argument is not a valid MySQL result resource in /home/fiestypics.com/www/cimages.php on line 30


Supplied argument is not a valid MySQL result resource in /home/fiestypics.com/www/cimages.php on line 48

Code is at that part just at above. It starts at line 30

[Defender1]

Yea yea, typo.

and check to make sure you're actually connecting to your database.