inserting checkbox values in mysql +PHP

Question

asadalim1 0 Light Poster

16 Years Ago

I am new to PHP or just not good at it lol, I have been trying to insert checkbox elements into a mysql database.

I have a table called Team with two collumns.

Id int not null auto increment PK
team varchar 30 not null

<?
    <body><form action="insert.php" method="post">

        select team:
        <input type="checkbox"  value="AR"    NAME="team[]">:Argentina
        <input type="checkbox"  value="PL"    NAME="team[]">:Spain
        <input type="checkbox"  value="GB"    NAME="team[]">:Holland
        <input type="checkbox"  value="FR"    NAME="team[]">:France
<br>

  <input name="confirm" type=submit id="confirm" style='border:1px solid #000000;font-family:Verdana, Arial, Helvetica, sans-serif ;font-size:9px;' value='Confirm'>
</body>
</form>
?>

This is the insert.php

<?php
$con = mysql_connect("localhost","****","*****");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
if (!$id) {
    $id= null;
}

<? 

 if(isset($_POST['team']))
{

$team = 1;

$insert=mysql_query("

INSERT INTO team('team') VALUES ('" . $team. ");

   ?>




Parse error: syntax error, unexpected '<' in C:\wamp\www\examples\insert.php on line 10

Any help or suggestions will be much appreciated.

mysql php

Edited 11 Years Ago by mike_2000_17 because: Fixed formatting

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Latest Post 16 Years Ago Latest Post by asadalim1

nav33n 472 Purple hazed!

16 Years Ago

<?php
if(isset($_POST['team']))
{
	foreach($_POST['team'] as $value){
		$insert=mysql_query("INSERT INTO team('team') VALUES ('$value')");
	}
}
?>
<html>
<body>
<form method="post" action="chk123.php">
<input type="checkbox" name="team[]" value="AG"> Argentina <br />
<input type="checkbox" name="team[]" value="GE"> Germany <br />
<input type="checkbox" name="team[]" value="BR"> Brazil <br />
<input type="submit" name="submit" value="submit">
</form>
</body>
</html>

There! That will work :)

Cheers,
Naveen

Edit: The error was because, you had opened php tag but you opened it again.

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ryan_vietnow 13 Posting Pro · Answer 1 · 2008-02-08T19:18:28+00:00

what's the line 10? I think you have left a bracket missing.

asadalim1 0 Light Poster · Answer 2 · 2008-02-08T20:52:11+00:00

asadalim1 0 Light Poster

16 Years Ago

Thank you :)

nav33n 472 Purple hazed! Team Colleague Featured Poster · Answer 3 · 2008-02-08T20:53:50+00:00

nav33n 472 Purple hazed!

16 Years Ago

:) You are welcome!

asadalim1 0 Light Poster · Answer 4 · 2008-02-08T22:04:28+00:00

<html>
<body>
<form method="post" action="chk123.php">
<input type="checkbox" name="team[]" value="AG"> Argentina <br />
<input type="checkbox" name="team[]" value="GE"> Germany <br />
<input type="checkbox" name="team[]" value="BR"> Brazil <br />
<input type="submit" name="submit" value="submit">
</form>
</body>
</html>




<?php
$con = mysql_connect("localhost","root","****");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());

if(isset($_POST['team']))
{   foreach($_POST['team'] as $value){  
    $insert=mysql_query("INSERT INTO team('team') 
  VALUES ('$value')");
    }
    }
    ?>

I am getting the error below, whats wrong??? :(

Parse error: syntax error, unexpected $end in C:\wamp\www\examples\chk123.php on line 13

nav33n 472 Purple hazed! Team Colleague Featured Poster · Answer 5 · 2008-02-08T22:06:47+00:00

nav33n 472 Purple hazed!

16 Years Ago

missing } after die function.

asadalim1 0 Light Poster · Answer 6 · 2008-02-08T23:53:35+00:00

It processes, but nothing is stored in the Mysql, table:(

nav33n 472 Purple hazed! Team Colleague Featured Poster · Answer 7 · 2008-02-09T00:04:29+00:00

$insert="INSERT INTO team('team')
VALUES ('$value')";
echo $insert;
mysql_query($insert);

Print your query. Execute the same in phpmyadmin or mysql. Check what's the error. Or you can give, mysql_query($insert) or die(mysql_error()); to know what is the error.

asadalim1 0 Light Poster · Answer 8 · 2008-02-11T18:05:24+00:00

Hi, i have tried this but doesn't return anything.

I'm Wondering whether my table design in mysql is incorrect.

I have only two collumns, one to store the ID and one to store the team.

Table Team 
[U]id[/U]      Int(11)    Not Null     auto_increment
team          varchar(30)   Not Null 




<?php
$con = mysql_connect("localhost","root","talk21");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
}
if(isset($_POST['team']))
{   foreach($_POST['team'] as $value){  
    $insert=mysql_query("INSERT INTO team('team') 
  VALUES ('$value')");
    }
    }

echo $insert;mysql_query($insert);

    ?>

nav33n 472 Purple hazed! Team Colleague Featured Poster · Answer 9 · 2008-02-11T18:12:24+00:00

<?php
$con = mysql_connect("localhost","root","talk21");
if (!$con) {
    die('Could not connect: ' . mysql_error());
}
if(isset($_POST['team'])) {    foreach($_POST['team'] as $value) {
    $insert="INSERT INTO team ('team') VALUES ('$value')";
echo $insert;
mysql_query($insert);
}
}

Try this. It will print out the query. Execute it in phpmyadmin or mysql. I don't think your query or the table structure is wrong. Maybe the values of $_POST isn't being passed. Well, execute these queries and find out why its not inserting !

asadalim1 0 Light Poster · Answer 10 · 2008-02-11T18:34:15+00:00

I tried the query it prints "INSERT INTO team ('team') VALUES ('BR')" correctly however it still doesnt store the value into the database, I even selected mysql_select_db("lastr", $con);$sql= the database still not working.

thank you for all your responses by da way.

<?php
$con = mysql_connect("localhost","root","talk21");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
if(isset($_POST))
{ foreach($_POST as $value) {
mysql_select_db("lastr", $con);$sql= $insert="INSERT INTO team ('team')
VALUES ('$value')";echo $insert;mysql_query($insert);
}
}

?>

nav33n 472 Purple hazed! Team Colleague Featured Poster · Answer 11 · 2008-02-11T19:33:27+00:00

umm.. You don't have a mysql_select_db in your script. If it isn't inserting, use die function with mysql_error with your query. For eg.

$query="select * from tablename";
mysql_query($query) or die(mysql_error());

So, if there's any problem with your query, it will output the error message, which you can use to debug your query.

asadalim1 0 Light Poster · Answer 12 · 2008-02-11T20:15:42+00:00

I have tried it with this function "mysql_query($query) or die(mysql_error());"

there is no error and nothing is still being inserted into the database :(.

<?php
$con = mysql_connect("localhost","root","talk21");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
if(isset($_POST))
{ foreach($_POST as $value) {
mysql_select_db("lastr", $con);$sql= $insert="INSERT INTO team ('team')
VALUES ('$value')";echo $insert;mysql_query($insert);
}
}

$query="select * from team";mysql_query($query) or die(mysql_error());

?>

nav33n 472 Purple hazed! Team Colleague Featured Poster · Answer 13 · 2008-02-11T20:18:36+00:00

Okay! Take out 'team' and put team. ie.,

$insert="INSERT INTO team (team)
VALUES ('$value')";

Tellme if it works.

asadalim1 0 Light Poster · Answer 14 · 2008-02-11T20:28:31+00:00

Still the same, Nothing going into the database, maybe should insert into id aswell.

<?php
$con = mysql_connect("localhost","root","talk21");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}

if(isset($_POST))
{ foreach($_POST as $value) {
mysql_select_db("lastr", $con);$sql= $insert="INSERT INTO team. ('team')
VALUES ('$value')";echo $insert;mysql_query($insert);
}
}

$query="select * from team";mysql_query($query) or die(mysql_error());
?>

nav33n 472 Purple hazed! Team Colleague Featured Poster · Answer 15 · 2008-02-11T20:30:57+00:00

Umm.. is Id an auto-increment field ? What did echo $insert print ? Could you show us the statement ? Did you execute that statement in phpmyadmin/mysql ?

asadalim1 0 Light Poster · Answer 16 · 2008-02-11T20:46:17+00:00

INSERT INTO team ('team') VALUES ('GE')INSERT INTO team ('team') VALUES ('BR')

the id is set to auto_increment, when i run the query on phpadmin this is the error.

#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''team') VALUES ('GE')INSERT INTO team ('team') VALUES ('BR')' at line 1

nav33n 472 Purple hazed! Team Colleague Featured Poster · Answer 17 · 2008-02-11T23:08:31+00:00

:) That's what I was talking about.

<?php
$con = mysql_connect("localhost","root","talk21");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("lastr");

if(isset($_POST['team']))
{ 
     foreach($_POST['team'] as $value) {
          $insert="INSERT INTO team (team) VALUES ('$value')";
     mysql_query($insert);
    }
}
?>

This will definitely work. :)

P.S: Next time you post your code, please post it within code-tags.

asadalim1 0 Light Poster · Answer 18 · 2008-02-12T16:57:36+00:00

Great, many thanks for your help :). say if i wanted to do a dynamic table in mysql to insert the checkbox values from is there any threads you could point me to for example or help.

nav33n 472 Purple hazed! Team Colleague Featured Poster · Answer 19 · 2008-02-12T18:35:10+00:00

I didn't get you. What do you mean by dynamic table ? you mean creating table on runtime ?

asadalim1 0 Light Poster · Answer 20 · 2008-02-12T20:59:41+00:00

I mean getting the checkbox values from a database and puting it on form instead or hardcoding it. e.g

<form name="form" method="GET" action=" <?php echo 'process.php' ?> ">
Navn<input type="text" name="Navn" /><br>
country<?php echo html_drop_down('select'); ?><br>

$sql = 'SELECT * From Countries';

while ($rec = mysql_fetch_array($result))
{
print "<input name=\"".$rec."\" type=\"checkbox\" value=\"".$rec."\" /> ".$rec
}

nav33n 472 Purple hazed! Team Colleague Featured Poster · Answer 21 · 2008-02-12T21:41:58+00:00

$query="select * from countries";
$result=mysql_query($query);
while($row=mysql_fetch_array($result)){
  $id=$row['id'];
 $country=$row['country'];
  print "<input type=\"checkbox\" name=\"chkbox[]\"  value=\"$id\">".$country."<br/>";
}

asadalim1 0 Light Poster · Answer 22 · 2008-02-12T23:42:36+00:00

I have this form code working it prints the checkoxes with the countries.

<form id="form1" name="form1" method="post" action="page2.php">
<?php
$sql = 'SELECT * From Countries';

$con = mysql_connect("localhost","root","talk21") or die;

$result = mysql_db_query(talk21, $sql);

while ($rec = mysql_fetch_array($result))
{
print "\n<input type='checkbox' name=Countries[]' value='".$rec."' > ".$rec." </input>";
}

?><br />
<input type="submit" name="Submit" value="Register" />
</label>
</form>

the page should process the form page2.php,

<?php
$countries = $_REQUEST["Countries"];

for(int $i = 0; $i < $countries.size; i++){
if(isset($countries[$i])){

}
}
?>

the code is incomplete

nav33n 472 Purple hazed! Team Colleague Featured Poster · Answer 23 · 2008-02-12T23:45:43+00:00

What is the problem ? And please post your code within CODE tags. Btw, there is no function called .size . Its count or sizeof.

asadalim1 0 Light Poster · Answer 24 · 2008-02-13T00:36:05+00:00

What is the problem ? And please post your code within code tags. Btw, there is no function called .size . Its count or sizeof.

yes you are right sizeof. ,the error is Parse error: syntax error, unexpected '}' in C:\wamp\www\examples\prew_akt.php on line 8.

I want to insert the values selected from the form into a table also, i want create a page where the values of the checkboxes can be edited CHANGED/DELETED.

nav33n 472 Purple hazed! Team Colleague Featured Poster · Answer 25 · 2008-02-13T09:31:38+00:00

I dont see any unexpected } in your code. Please post your latest code(with code tags).. And the idea (what I follow) to edit the check box's value is, When the user is in the edit screen and he has changed the checkbox's selection, I first delete all the previously selected checkboxes for that particular user and then re-insert the new values again.

asadalim1 0 Light Poster · Answer 26 · 2008-02-13T19:01:29+00:00

Hi again yes I want one page to edit the checkboxs values so they can be updated or added/deleted.

the code here is to store the selected values into a database after the form is submitted.

 <?php
$countries = $_REQUEST["Countries"];

for(int $i = 0;  $i < $countries.sizeof.; i++){
if(isset($countries[$i])){

}
}

This code below is intended for to update or add/delete countries.

// ADDING Countries HERE //
if(array_key_exists('add', $_GET)){
    $id = $_POST['id'];
    $countries = $_POST['countries'];

    $sql = mysql_query("INSERT INTO countries (id, countries,)
          VALUES('$id', '$countries', )") or die (mysql_error());
}
    if (!$sql) {
        $message = "Error has occured";

        } else {
        $statusa = '1';
        $message = "editcountry has been ADDED successfully";
    }
// START EDITING countries //

if(array_key_exists('edit', $_GET)){
    $id = $_POST['editid'];
    $editcountry = $_POST['editcountry'];

    $sqla = "UPDATE `countries` SET `editcountry` = '$editcountry',WHERE `id` = '$id' ";

    $sql = mysql_query($sqla) or die (mysql_error());
    $statusa = '1';
    $message = "countries has been EDITED successfully";

?>
<html>
<body>

      Add Countries
        <form action="admin.php?add=1" method="post" name="add" id="add">

        <input type='checkbox' name=Countries[]' value='".$rec['id']."' > ".$rec['Country']." </input>";     
        <input name="submit" type="submit" id="submit" value="Submit">
        </form>
      <p>

      <?
      if ($statusa == 1){
        echo $message ;
      }
      ?>

      Edit Countries
      <?
        $sql = "SELECT * FROM `property`";
        $send = mysql_query($sql) or die(mysql_error());

        ?>
            <form action="admin.php?edit=1" method="post" name="edit" id="edit">
        ID 
        <select name="editid" id="editid">
        <?  while ($row = mysql_fetch_object($send)) {
            echo "<option>$row->id<option>";
            }
        ?>
        </select>        
        <br>
        DELETE 
        <select name="editdelete" id="editdelete">
          <option selected>No</option>
          <option>Yes</option>
          </select>
        <br>

       <input type='checkbox' name=editCountries[]' value='".$rec['id']."' > ".$rec['Country']." </input>";  

</body>
</html>

nav33n 472 Purple hazed! Team Colleague Featured Poster · Answer 27 · 2008-02-13T22:52:41+00:00

Error 1.

for(int $i = 0; $i < $countries.sizeof.; i++){

My apologies. I said there is no sizeof. But there is sizeof. But this isn't the way to use it. Its sizeof($variable);
Error 2.

$sql = mysql_query("INSERT INTO countries (id, countries,)
VALUES('$id', '$countries', )") or die (mysql_error());

Wrong query. What's the "," doing ?
Error 3. You are using method=POST in the form . But you are searching in $_GET ?

if(array_key_exists('edit', $_GET)){

nav33n 472 Purple hazed! Team Colleague Featured Poster · Answer 28 · 2008-02-13T23:39:35+00:00

Anyway, here is a simple example to insert teams to the table, list selected teams and delete the teams from the table.

<?php //insert to the table
if(isset($_POST['submit'])){
	mysql_connect("localhost","root");
	mysql_select_db("test");
	foreach($_POST['country'] as $value){
		$query="insert into country (team) values ('$value')";
		mysql_query($query);
	}
}
?>
<html>
<body>
<form method="post" action="checkbox.php">
<input type="checkbox" name="country[]" value="AG">Argentina <br />
<input type="checkbox" name="country[]" value="GE">Germany <br />
<input type="checkbox" name="country[]" value="BR">Brazil <br />
<input type="checkbox" name="country[]" value="SW">Sweden <br />
<input type="checkbox" name="country[]" value="EN">England <br />
<input type="checkbox" name="country[]" value="SC">Some country <br />
<input type="submit" name="submit" value="submit">
</form>
</body>
</html>

<?php //list inserted teams select teams you want to delete and click submit and it will delete. 
	mysql_connect("localhost","root");
	mysql_select_db("test");
	
	if(isset($_POST['submit'])) {
		foreach($_POST['country'] as $value) {
			$query="delete from country where team='$value'";
			mysql_query($query);
		}
	}
	
	$query="select * from country";
	$result=mysql_query($query);
	echo "<form method=\"post\" action=\"editcheckbox.php\">";
	while($row=mysql_fetch_array($result)) {
		$country=$row['team'];
		echo "<input type=\"checkbox\" name=\"country[]\" value='$country'>$country<br />";
	}
	echo "<input type='submit' name='submit' value='submit'>";
	echo "</form>";
?>

I hope that helps :)