Table data from db as new form input

[tomycon]

Hi all,

Ive got a page with code that generates a table populated with values from a postgresql db. My code looks like this:

PHP Code:

    $dbconn = pg_Connect ("dbname=bkadata user=httpd password=httpd");

    if (!$dbconn) {
             echo "</table>An error occurred.\n";
             exit;
    }

    $result = pg_Exec($dbconn,
             "SELECT joblog.job_no, joblog.job_name, joblog.status, joblog.bill_staff
                FROM joblog
                WHERE job_no >= 205000 AND job_no < 300000
                ORDER BY job_no DESC;");

    if (!$result) {
             echo "</table>An error occurred.\n";
             exit;
    }

    $num = pg_NumRows($result);
    $i = 0;

    while ($i < $num) {
            if ($i % 2 ==0) {
                $bgcolor='#f3f3f3';
                } else {
                $bgcolor='#e3e3e3';
                }
            echo "<TR bgcolor=$bgcolor><TD>";
            echo pg_Result($result, $i, "job_no");
            echo "</TD><TD>";
            echo pg_Result($result, $i, "job_name");
            echo "</TD><TD>";
            echo pg_Result($result, $i, "status");
            echo "</TD><TD>";
      echo pg_Result($result, $i, "bill_staff");
      echo "</TD></TR>";
            $i++;
    }

    pg_FreeResult($result);
    pg_Close($dbconn); 


This works fine. Now I want to make the resulting table a form, with a submit icon in each row that will pass a value from the particular row chosen when a user clicks on the icon. The submit takes the user to another page where more info about the row chosen is displayed in another table generated from a new query to a different db.

So far I've got this code:

PHP Code:

$dbconn = pg_Connect ("dbname=bkadata user=httpd password=httpd");

    if (!$dbconn) {
             echo "</table>An error occurred.\n";
             exit;
    }

    $result = pg_Exec($dbconn,
             "SELECT joblog.job_no, joblog.job_name, joblog.status, joblog.bill_staff
                FROM joblog
                WHERE job_no >= 205000 AND job_no < 300000
                ORDER BY job_no DESC;");

    if (!$result) {
             echo "</table>An error occurred.\n";
             exit;
    }

    $num = pg_NumRows($result);
    $i = 0;

    while ($i < $num) {
            if ($i % 2 ==0) {
                $bgcolor='#f3f3f3';
                } else {
                $bgcolor='#e3e3e3';
                }
                $jobno[] = pg_Result($result, $i, "job_no");

            echo "<TR bgcolor=$bgcolor><TD>";
            echo "<INPUT TYPE=hidden  name=cntct_jno value=";
            echo $jobno[$i];
            echo ">";
            echo $jobno[$i];
            echo "</imput></TD><TD>";
            echo "<input type=image src=contactb2.gif border=0 width=24 height=24 alt=contactb2.gif value=submit>";
            echo "</TD><TD>";
            echo pg_Result($result, $i, "job_name");
            echo "</TD><TD>";
            echo pg_Result($result, $i, "status");
            echo "</TD><TD>";
            echo pg_Result($result, $i, "bill_staff");
            echo "</TD></TR>";
            $i++;
    }

    pg_FreeResult($result);
    pg_Close($dbconn); 


The above code displays the initial table with the new submit icon fine. But when I click on the icon for any row, it always passes the value for the last row in the table only!

I've tried several variations on the above code usually with the same result. What am I missing here?

Any help is appreciated!

Thanks!

[sde]

the first thing i see is the closing input tag:

PHP Code:

echo "</imput></TD><TD>"; 


the problem is with the way you are forming your html though. the submit button, submits the entire form ( not just the last hidden input )

you could either make a new form for each table row, or just send the job id in the URL.

the first option, your html row would look like this:

PHP Code:

while ($i < $num) {
            if ($i % 2 ==0) {
                $bgcolor='#f3f3f3';
                } else {
                $bgcolor='#e3e3e3';
                }
                $jobno[] = pg_Result($result, $i, "job_no");

            echo "<form method=POST action=...><TR bgcolor=$bgcolor><TD>";
            echo "<INPUT TYPE=hidden  name=cntct_jno value=";
            echo $jobno[$i];
            echo ">";
            echo $jobno[$i];
            echo "</imput></TD><TD>";
            echo "<input type=image src=contactb2.gif border=0 width=24 height=24 alt=contactb2.gif value=submit>";
            echo "</TD><TD>";
            echo pg_Result($result, $i, "job_name");
            echo "</TD><TD>";
            echo pg_Result($result, $i, "status");
            echo "</TD><TD>";
            echo pg_Result($result, $i, "bill_staff");
            echo "</TD></TR></form>";
            $i++;
    } 


or just pass the job variable in the url and make the image a link.

PHP Code:

while ($i < $num) {
            if ($i % 2 ==0) {
                $bgcolor='#f3f3f3';
                } else {
                $bgcolor='#e3e3e3';
                }
                $jobno[] = pg_Result($result, $i, "job_no");

            echo "<TR bgcolor=$bgcolor><TD>";
            echo $jobno[$i];
            echo "</imput></TD><TD>";
            echo "<a href=\"nextpage?cntct_jno={$jobno[$i]}\"><img  src=contactb2.gif border=0 width=24 height=24 alt=contactb2.gif border=0></a>";
            echo "</TD><TD>";
            echo pg_Result($result, $i, "job_name");
            echo "</TD><TD>";
            echo pg_Result($result, $i, "status");
            echo "</TD><TD>";
            echo pg_Result($result, $i, "bill_staff");
            echo "</TD></TR>";
            $i++; 


the idea is that you just send the job number in the url with a normal link around your icon. i.e. http://mysite.com/nextpage.php?cntct_jno=15

then, in the next page, you use use $_GET['cntct_jno'] to get the variable.

[tomycon]

Thank You!

Very helpful and informative too! I'm fairly new at this and had not come across the $_GET['cntct_jno'] function. That alone will solve some other issues I'll be getting into!