swatisonee Posted November 25, 2010 Share Posted November 25, 2010 I continue to have problems with this issue - primarily relates to the the placement of the code for processing the 2nd form on a page. This is what i wish to do: a. Form 1 - Select a User b. For selected user, list all the records c. Form 2 - Choose 1 record from the list. d. Update the record. Here's what i have so far. I've just put the structure out here as the error comes up when form 2 tries to get processed . I dont know if the issue is with the way i 've named the forms or the submit buttons. All pointers appreciated. Many thanks. Swati <? (processing form1) =================== if($_POST['submit-user']){ $user = $_POST["user"]; $result = mysql_query("SELECT ..."); if ($myrow = mysql_fetch_array($result)) { do { $xid=$myrow["Xid"]; //this has to be form 2 form2 ===== <form method="post" action="form2.php"> // is this correct - the first form was a "self" printf("<tr><td><input type=\"radio\" name=\"choice\" value=%d><td> <td>%s </td> <td>%s </td> </tr></p></table>", $xid , $myrow["timein"], $myrow["trip"] ); } while ($myrow = mysql_fetch_array($result)); } echo "</select>\n"; } <input type="submit" name="submit-data" value="Submit"> </p> </form> (end of form 2) processing form 2 - but where does it get placed ? ================== if($_POST['submit-data']){ $sscid = $_POST["choice"]; $result11 = mysql_query("SELECT * FROM `Table` WHERE `SSCID` = $sscid "); $myrow11 = mysql_fetch_array($result11); process this data } // does the bracket close here or below with the bracket of form 1 ?> // end processing of both forms. <html> <head> </head> <?php } else { ?> form1 ===== <FORM METHOD=post ACTION="<? echo $PHP_SELF ?>"> <select name="user"><option value="">[select One] <? $result = mysql_query("SELECT ..."); if ($myrow = mysql_fetch_array($result)) { do { $userid=$myrow["Userid"]; printf("<option value = '$userid'> %s", $myrow["Username"]); } while ($myrow = mysql_fetch_array($result)); echo "</select>\n"; } ?> <input type="submit" name="submit-user" value="Submit"> </p> </form> (end of form 1) <?php } ?> </html> Quote Link to comment Share on other sites More sharing options...
trq Posted November 25, 2010 Share Posted November 25, 2010 Is this all in one file? A forms action should point to the file that is going to do the processing. [ot] $PHP_SELF has long been depricated and should no longer be used. If you want your form to post bake to itself, just leave the action blank. eg; ection="" Do not use <? short tags. Use <?php always. [/ot] Quote Link to comment Share on other sites More sharing options...
swatisonee Posted November 25, 2010 Author Share Posted November 25, 2010 hi: yes and thats where i know i'm making a mistake but dont know how to correct it : a. If Form 1 = <FORM METHOD=post ACTION="<? echo $PHP_SELF ?>"> (noted your comment on php self.- thanks) then its result - that is the choice of user will go into Form 2 b. Form 2 = <form method="post" action="form2.php"> will display all the records for that user and then one of the records will be chosen for further processing which is basically updating the data into a mysql table . thanks Quote Link to comment Share on other sites More sharing options...
litebearer Posted November 25, 2010 Share Posted November 25, 2010 Perhaps along these lines.... if !isset $_POST['$process'] display form 1 hidden field $process = 1 post to self if isset $_POST['$process'] AND $_POST['$process']= 1 display form 2 hidden field $process = 2 post to self if isset $_POST['$process'] AND $_POST['$process']= 2 do your updating or whatever Quote Link to comment Share on other sites More sharing options...
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