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Categorieën > PHP & SQL

submit werkt niet (Opgelost)

pim1991 - 23/07/2007 17:08 (laatste wijziging 23/07/2007 17:09)

PHP interesse

Hallo,

Ik heb 2 formulieren in 1 pagina staan.
1 formulier werkt wel maar de andere niet.

Code verwerking die het wel doet:

php code - Bekijk de code zonder highlighting - Klap code in

  if($_POST['submit'] && $_POST['status']){
  $user_id = $_POST['user_id'];
  $status = $_POST['status'];
  $war_id = $_POST['war_id'];

  $sql = "INSERT INTO wars_inschrijven (user_id, status, war_id) VALUES ('".$user_id."', '".$status."', '".$war_id."')";
  mysql_query($sql);
  }

if($_POST['submit'] && $_POST['status']){
  $user_id = $_POST['user_id'];
  $status = $_POST['status'];
  $war_id = $_POST['war_id'];
 
  $sql = "INSERT INTO wars_inschrijven (user_id, status, war_id) VALUES ('".$user_id."', '".$status."', '".$war_id."')";
  mysql_query($sql);
  }

code verwerking die het niet doet:

php code - Bekijk de code zonder highlighting - Klap code in

if($_POST['submit2'] && $_POST['text']){
  $posted_by = $_POST['posted_by'];
  $text = $_POST['text'];
  $warid = $_POST['warid'];

  $sql = "INSERT INTO wars_reactie (posted_by, text, warid) VALUES ('".$posted_by."', '".$text."', '".$warid."')";
  mysql_query($sql);
  }

if($_POST['submit2'] && $_POST['text']){
  $posted_by = $_POST['posted_by'];
  $text = $_POST['text'];
  $warid = $_POST['warid'];
 
  $sql = "INSERT INTO wars_reactie (posted_by, text, warid) VALUES ('".$posted_by."', '".$text."', '".$warid."')";
  mysql_query($sql);
  }

code niet werkende form

php code - Bekijk de code zonder highlighting - Klap code in

echo "<form method=post acion=".$PHP_SELF."><center><textarea cols=40 rows=5 name=text></textarea><input type=hidden value=".$userdata['user_id']." name=posted_by><input type=hidden value=".$id." name=warid><br /><input type=submit value=Post reactie! name=submit2></center></form><br />";

echo "<form method=post acion=".$PHP_SELF."><center><textarea cols=40 rows=5 name=text></textarea><input type=hidden value=".$userdata['user_id']." name=posted_by><input type=hidden value=".$id." name=warid><br /><input type=submit value=Post reactie! name=submit2></center></form><br />";

code werkende form:

php code - Bekijk de code zonder highlighting - Klap code in

echo "<form method=post action=".$PHP_SELF.">";
echo "<tr><td width=125>Kom je?</td><td>";
echo "<select name=\"status\">
<option value=\"Ja!\">Ja</option>
<option value=\"Misschien\">Misschien</option>
<option value=\"Nee\">Nee</option>
</select>";
echo "<input type=hidden name=user_id value=".$userdata['user_id']."><input type=hidden name=war_id value=".$id."></td></tr>";
echo "<tr><td width=125>&nbsp;</td><td><input type=submit name=submit value=Oke></form></td></tr>";

echo "<form method=post action=".$PHP_SELF.">";
echo "<tr><td width=125>Kom je?</td><td>";
echo "<select name=\"status\">
<option value=\"Ja!\">Ja</option>
<option value=\"Misschien\">Misschien</option>
<option value=\"Nee\">Nee</option>
</select>";
echo "<input type=hidden name=user_id value=".$userdata['user_id']."><input type=hidden name=war_id value=".$id."></td></tr>";
echo "<tr><td width=125>&nbsp;</td><td><input type=submit name=submit value=Oke></form></td></tr>";

Alvast bedankt voor de help!

4 antwoorden

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NTS64 - 23/07/2007 17:14

PHP gevorderde

php code - Bekijk de code zonder highlighting - Klap code in

echo "<form method=post acion=".$PHP_SELF."><center><textarea cols=40 rows=5 name=text></textarea><input type=hidden value=".$userdata['user_id']." name=posted_by><input type=hidden value=".$id." name=warid><br /><input type=submit value=Post reactie! name=submit2></center></form><br />";

echo "<form method=post acion=".$PHP_SELF."><center><textarea cols=40 rows=5 name=text></textarea><input type=hidden value=".$userdata['user_id']." name=posted_by><input type=hidden value=".$id." name=warid><br /><input type=submit value=Post reactie! name=submit2></center></form><br />";

Er staat acion ipv action

Koen - 23/07/2007 17:16 (laatste wijziging 23/07/2007 17:16)

PHP expert

php code - Bekijk de code zonder highlighting - Klap code in

if(IsSet($_POST['submit']) && $_POST['status']){
$user_id = $_POST['user_id'];
$status = $_POST['status'];
$war_id = $_POST['war_id'];

$sql = "INSERT INTO wars_inschrijven (user_id, status, war_id) VALUES ('".$user_id."', '".$status."', '".$war_id."')";
mysql_query($sql);
} elseif(IsSet($_POST['submit2']) && $_POST['text']) {
$posted_by = $_POST['posted_by'];
$text = $_POST['text'];
$warid = $_POST['warid'];

$sql = "INSERT INTO wars_reactie (posted_by, text, warid) VALUES ('".$posted_by."', '".$text."', '".$warid."')";
mysql_query($sql);
} else {
// de form
}

if(IsSet($_POST['submit']) && $_POST['status']){
$user_id = $_POST['user_id'];
$status = $_POST['status'];
$war_id = $_POST['war_id'];
 
$sql = "INSERT INTO wars_inschrijven (user_id, status, war_id) VALUES ('".$user_id."', '".$status."', '".$war_id."')";
mysql_query($sql);
} elseif(IsSet($_POST['submit2']) && $_POST['text']) {
$posted_by = $_POST['posted_by'];
$text = $_POST['text'];
$warid = $_POST['warid'];
 
$sql = "INSERT INTO wars_reactie (posted_by, text, warid) VALUES ('".$posted_by."', '".$text."', '".$warid."')";
mysql_query($sql);
} else {
// de form
}

edit: text vergeten xD
kan je ook doen.

oh boven was me te vlug af, had de acion niet eens gezien xD

pim1991 - 23/07/2007 17:50
PHP interesse		Werkt allebij niet

barone - 28/07/2007 13:55
Onbekend

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