can I pass mysql_result() as a hidden form item

mmoranuk

I have a result vaiaible $result which is filled using:

$result = mysql_query($query);

I want to use this on two different web pages. Rather than rerun the query, is it possible to pass the $result as a hidden item on a form on the first page to the second using:

<input type=\"hidden\" name=\"questions_result\" value=\"$result\">

I have tried it and find there is no value in $result when I try to access it using:

$question_id = mysql_result($questions_result,$i,"question_id");

Any ideas or alternative suggestions would be great.

I would actually like to put it into a new array on the second page so that I can add a new set of values to each row (depending on if it was 'checked' on the first page).

Thanks in advance. Mark.

Roger_Ramjet

You will need to extract the data from $result using mysql_fetch_array.

You can also save the array as a session var so that it is there from page to page, instead of putting it into a hidden form element.

Mysql uses query cacheing so that rerunning the query has very little overhead. Infact, if this query is static in your code, and the underlying tables do not change, then you are probably getting the results from cache in the first place.

mmoranuk

Thanks for the response....I tried this but it doesn't work. the first value ($num_questions) is also a hidden field and it comes through fine. The while loop doesn't run though (ie, no rows).

echo "the question count is: $num_questions."; 
WHILE ($row = mysql_fetch_array($questions_result, MYSQL_NUM)) {
		echo "The value for question $row[0] is... <br>";
	}

I'd rather not rerun the query - loose code as the data in the table could change. I will wrap the important bits up in transactions.

Sessions may be something I look at for the future but I'd like the quick fix for this using hidden fields.

Any obvious ideas?

Roger_Ramjet

Well, you've answered your own question: your query is not returning any results. Test the num rows for the query resultset first, then echo or printf() extracted data.

$result is a resource that WILL NOT display its contents. If you try to echo $result all you will get is the resource ID.

mmoranuk

but there are rows - if I test the row_count passed through from the original page I get a correct count. If I try to count using the $result once it is passed through I get no rows.

This leads me to think that I'm no longer holding any resultset. The data/query hasn't changed - I just can't access $result.

I'm not trying to display $result, I'm trying to do a mysql_fetch_array() on it as you originally suggested.

Roger_Ramjet

No one said anything about passing $result, did I not make myself clear?😕

THAT WILL NOT WORK

test and echo on the page that runs the query. YOU CAN NOT PASS $result . I said that in my first answer.

You will need to extract the data from $result using mysql_fetch_array .

You can also save the array as a session var so that it is there from page to page, instead of putting it (ie the extracted array) into a hidden form element.

To make the query results available to php you use mysql_fetch_array on the page that ran the query. You then put the array into the hiden field. You can then use them from the second page.

Really, take the time to read the manual and learn the basics first.

mmoranuk

I don't think you did make it clear in the first response.
I can't find anywhere obvious in a manual that says you can't pass $result so "reading the manual and learning the basics" wouldn't help here.

I do a lot of database work but am fairly new to PHP so don't treat me like an idiot.

I'm just trying to write some code that works - please don't get annoyed with me unless I'm taking the pi55.

I understand your answer this time and I'll go away and try to make it work - thanks for your time.

Roger_Ramjet

from the manual

mysql query

Return Values

For SELECT, SHOW, DESCRIBE or EXPLAIN statements, mysql_query() returns a resource on success, and FALSE on error.

For other type of SQL statements, UPDATE, DELETE, DROP, etc, mysql_query() returns TRUE on success and FALSE on error.

The returned result resource should be passed to mysql_fetch_array(), and other functions for dealing with result tables, to access the returned data.

resource
Resource

A resource is a special variable, holding a reference to an external resource. Resources are created and used by special functions. See the appendix for a listing of all these functions and the corresponding resource types.

And NO, I'm not an expert in PHP, which is why I know where to find this in the manual - I have to refer to it every time I write new code.

mmoranuk

Thanks for that - I've read the page many times before. I still can't see where it says "don't pass a resource between two web pages" - which was what I was trying to do.

It seems obvious now that the resource reference is broken when trying to pass it. I didn't know this as I've never tried it before. I was hoping someone could explain why not or offer an alternative. I'm going to try your suggestion of building an array and passing that instead. It would have been nice to be able to pass the resource though.... :-)

Roger_Ramjet

It is "a reference to an external resource", ie a pointer to the resource, not the resource itself: passed by reference not by value. When the thread that created it terminates, ie when your script finishes, the memory the thread and associated resources used is freed for other processes and threads. The resource ceases to exist and so the ref is no longer valid.