Help Needed

lanesharon

I have a contact form that utilizes a table of questions and answers, instead of a visual confirmation. They are stored in a mySQL db, in ISAM id, order. For each one there is a question and answer in the table:

1 question1 answer1
2 question2 answer2
3 question3 answer3
etc.....

I open the database, then read the table with this code:

//$rs = mysql_query("select question, answer from question_table order by rand() limit 1");
//$row = mysql_fetch_array($rs);
$rs = mysql_query("select * from question_table order by rand() limit 1");
$row = mysql_fetch_array($rs);
$randomq=$row[question];
$randoma=$row[answer];

echo "rs=";
echo $rs;
echo "<br>";
echo "row=";
echo $row;
echo "<br>";
echo "randomq=";
echo $randomq;
echo "<br>";
echo "randoma=";
echo $randoma;

Throughout the process, I use these same echoes to try to see where the problem is.

My result for $rs is always rs=Resource id #4
My result for $row is always row=Array
My result for randomq & randoma are questions and answers in the table. I know they are random because when I refresh, they change. So, they are showing up okay. My problem is when I click on the Submit button for the contact input form, they change. So, they kick out an error message in which I have embedded echoes to displa them. I cannot figure out how they change, since the only other code that even utilizes them is the original code that displays the question (which displays fine) and asks the users to input the answer; and the echoes to put them out in an error message, if there is an error. The change is random in order (not the next one in the array or the last one in the array). But, it is always the correct answer for the correct question, so I know it is the entire row that is changing.

I am really new at php, so I am hoping I am just being stupid and forgetting something obvious.

Thanks for any help you can give me. Take Care, Sharon

YAOMK

are you posting to the same script? if so how are you preventing the query from running again? and more importantly how are you passing the values you got in your first result set?

lanesharon

The mysql statement I listed above are in a config file. The post statements are in a form file. The form file calls in the config file (require_once).

The question/answer values are assigned in the config file. The entered answer is assigned by a $_Post in the form file. I know that the passing of the values works okay, because I commented out the database call and assignments and 'hard coded' one question and answer. Then I got no errors. Once I use the table, with the multiple rows, I get an error.

I know I am missing something very basic here, but as I said, I am a newbie.

YAOMK

Please provide us with the error message you are getting. What I wanted to know was whether the form was posting to itself or another script. If you are not preventing your query from running more than once then that would change the values in the result set and therefore may not match the previous one

self.php

<?php
echo '<FORM METHOD="post" ACTION="self.php">';
?>

lanesharon

<form id="form" method="post" action="<?php echo(''.htmlentities($_SERVER["PHP_SELF"]).''); ?>#results">

#results is the success or error messages.

I'll modify the line and hard-code the url to test it out. The error message is part of the coding and has to do with the answer to the question not equaling the input answer. But, that only happens when it processes the form. As I said, I have put echoes in to track the process. If you want, you can try it and see what I mean here by looking at this test:
http://rare-cancer.org/test/form.php

Thanks so much for trying to help me. Take Care, Sharon

YAOMK

Yeap $randomq and $randoma are changing because you are re-running your query and since your query selects in a random order the order of the result set gets switched around. You need to use a condition on the call to the function that runs your query.

if(!isset($_POST['submit']))
{
//call function that runs query.....
}
//continue execution .....

You also need to pass the result of $randomq and $randoma to the script, you could do this by using hidden fields:

echo "<input type='hidden' name='randomq' value='$randomq'>";
$randomq = $_POST['randomq']['1'];
//..........

As far as the $randoma, you could store it temporarily in the database, but then you would need to get fancier with your code and implement sessions to be able to identify the client ant relate it with the appropriate value for $randoma. I wouldn't just pass it in a hidden field or query string as it defeats the purpose of your validation. Perhaps someone else may have a better solution.

lanesharon

I am a little confused by your answer. In essence, randoma is from the database. It is part of the question_table I am randomly drawing from. And, I only have one query statement in config.php; which is called in by form.php at the very beginning of it. And that is this code:

$rs = mysql_query("select * from question_table order by rand() limit 1");
$row = mysql_fetch_array($rs);
$randomq=$row[question];
$randoma=$row[answer];

Can you explain to me (a newbie) how the query is getting re-run? Thanks.

YAOMK

Sure, when you go to yoursite.com/form.php the form runs for the first time setting all the values, then when you click that submit button, you reload form.php and the script runs yet again, but this time $randomq and $randoma are getting their values in a different order as they did before, so to retain the older values, you need a way to stop the query from running and a way to have the older values (the ones assigned by your script the first time it ran) available to the present state of the script.

Look at it this way: in the firs state of your script (before the form is submitted), you ask your user for input, then in the second state of your script you are validating that data, therefore you need all the data from the first state of the script available in the second state of it.