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Gavski

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  1. Well, I now have value $box2[0] successfully stored as a $var. this will allow me to finish building the page and others that use similar. seems my mistake was thinking $new_var = $_POST['box2']; made $new_var as a string but it's not, it's an array that can then be imploded. I have learned a lot from this, but most of all how good ppl will hep you out if you get stuck. so many cheers for all the help and thanks for your patience.
  2. Great, nearly there? page source now shows $id values, printf('<pre>%s</pre>', print_r($_POST, true)); returns - Array ( [box2] => Array ( [0] => 5 ) I now have the value i want in box2 array. Final problem I cant seem to the value out of box2 array? eg $new_var = $box2[0]; echo $new_var; doesnt work, also echo count($box2); returns 0 I really need to get $box2[0] as a var, I've tried implode() but get invalid parameter error??? thanks
  3. no, the checkbox doesn't display a value but the displayed list does display a value for $id, therefore $id definitely does have a value but the checkbox doesn't seem to wan to post it, this is my problem. cheers and thanks for your patience
  4. also, If I substitute $id in the 'value' of the checkbox with a static value( eg 1,2 or A,B) box2[ ] posts this value fine.
  5. sorry, I simplified the form to post it on here and the ,"] were the end of another set of values. The check box's do appear ok, one along side each record that is displayed but when I tick a check box the box2[ ] array fails to post the 'value' $id.
  6. This doesn't work?? when a checkbox is checked, and the form posted, the posted value $_POST['box2'] is null, I am expecting it to hold the value $id for the record associated wit it?
  7. I'm trying to display an array in HTML and next to each record is a checkbox, as below. I then want to be able to select a record by checking the checkbox and posting the 'value' of box2[ ] for that record. The problem I am having is the checkbox doesn't seem to post the value ($id), it only seems to handle static values (1,2,3 etc) instead of a string. Any ideas how to work this, or suggesions for marking an individual record to then post it's id? thanks $linkid=dbconnect("db"); mysql_select_db("db"); if(!($result = mysql_query("SELECT * FROM table WHERE id = '$id'"))) die ("Result error"); { echo "<table>\n"; while ($myrow = mysql_fetch_array($result)) { printf("<tr><td>%u</td><td><input type='checkbox' name='box2[ ]' value='$id'></td><td>%s</td><td>%s</td></tr>\n", $myrow["id"],$myrow["descr"],$myrow["batno"],"]); } echo "</table>\n"; }//endwhile }//endif
  8. Thanks Cyber Robot, Managed to do this from your help and it works fine. $num_results = mysql_num_rows($result); if($num_results > 0) returns 0 if no records found. Also Barand, I have gone through my function and amended it (I kind of knew globals were a bad thing), now running like a well oiled address book, thanks for all help and tips
  9. This is a function that searches an address DB, compares the $name with 'LIKE' and returns the address details. If no address is found, I want to go to another function. In my form, if I leave $name blank it works, but if I put a $name that is not in the database, $result still returns a value and doesn't seem to recognize that there is no record?? any ideas, thanks. <?php function getsupp() { global $name,$address1,$address2,$address3,$address4,$pcode,$email; mysql_select_db("picture_cards"); $result = mysql_query("SELECT * FROM supplier WHERE name LIKE('$name%')") or die(mysql_error()); if($result ){ while($row = mysql_fetch_array($result)) { $name = $row['name']; $address1 = $row['address1']; $address2 = $row['address2']; $address3 = $row['address3']; $address4 = $row['address4']; $pcode = $row['pcode']; $email = $row['email']; }//endwhile }//endif }// endfunc getsupp() ?>
  10. Thanks, that's great, Ive found a similar but simpler solution - if ($_SERVER["REQUEST_METHOD"] == "POST") { myfunction(); } test whether a post has been made (in pre hypertext process.) thanks again.
  11. Sorry here it is again without the glaring error - <?php session_start(); $_SESSION['regName'] = $regValue; $token = 0; //here's the function function doThis() { echo "Do this Function";} if ($token = 1) { doThis(); } $token = 1; echo $token; ?> <form method="post" action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]);?>" > <input type="text" name="regName" value=""> <input type="submit"> </form>
  12. Here's my form, it contains a function (really!). I want this function to work only after the form has been posted (not on initial loading). Here PHP reads the value of $token inside the function on loading (setsit to 1) and so executes the function when the page is loaded, there must be a way round this? <?php session_start(); $_SESSION['regName'] = $regValue; $token = 0; //here's the function function doThis() { echo "Do this Function";} if ($token = 1) { doThis(); $token = 1; } echo $token; ?> <form method="post" action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]);?>" > <input type="text" name="regName" value=""> <input type="submit"> </form>
  13. I have a form that contains a function. i want this function to be activated only when the user reposts the page to itself. (php_self). ie. the user access the page and the function is hidden, the user then inputs data, repost the page and the function is activated. I've tried flagging the function with a variable - $var = 0, then $var = 1 at the end of the page but $var just resets itself to 0 when the page is posted. Sure this is simple but I just cant get it? thanks
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