[RESOLVED] foreach problem

qphqp

i have a code like this called page.php


<html>
<head>
<title>Untitled Document</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
</head>

<body>
<form action="save.php" method="post">
<table width="100%" height="550">
<tr height="225" align="center">
<td width="50%" align="center"><select name="menu[]">
  <option value="www.xxx.com">xxx</option>
  <option value="www.yyy.com">yyy</option>
  <option value="www.zzz.com">zzz</option>
  <option value="www.ttt.com">ttt</option>
</select>
</td>
<td width="50%" align="center"><select name="menu[]">
  <option value="www.xxx.com">xxx</option>
  <option value="www.yyy.com">yyy</option>
  <option value="www.zzz.com">zzz</option>
  <option value="www.ttt.com">ttt</option>
</select>
</td>
</tr>
<tr height="225" align="center">
<td width="50%" align="center"><select name="menu[]">
  <option value="www.xxx.com">xxx</option>
  <option value="www.yyy.com">yyy</option>
  <option value="www.zzz.com">zzz</option>
  <option value="www.ttt.com">ttt</option>
</select>
</td>
<td width="50%" align="center"><select name="menu[]">
  <option value="www.xxx.com">xxx</option>
  <option value="www.yyy.com">yyy</option>
  <option value="www.zzz.com">zzz</option>
  <option value="www.ttt.com">ttt</option>
</select>
</td>
</tr>
</table>
<input type="submit" value="send" name="B1">
</form>
</body>
</html>

and this goes to save.php

<?php
	@mysql_connect("localhost", "root", "")
  or die ("mistake!");
 @mysql_select_db("url")
  or die("mistake!");

foreach(??){
?
?
} 
mysql_close();
?>

and i want get a result in db table

id url1 url2 url3 url4
1 www.zzz.com www.xxx.com www.yyy.com www.ttt.com
2 .... ... .... ....
each of url1 url2 url3 url4 will contain selected option from menu.

What can be the code in foreach to get this result?i have try so many times but i couldn't.

paulnaj

Hi there,

Just to make sure I understand ...

your table has five fields. An id field (I'm assuming it's autoincrement) and at least 4 other fields that each contain urls, named url1 thru url4.

Your SQL statement should look something like:

<?php
$sql = "
INSERT INTO `table` (`id`, `url1`, `url2`, `url3`, `url4`) 
VALUES (
    NULL
,    '".$_POST['menu'][0]."'
,    '".$_POST['menu'][1]."'
,    '".$_POST['menu'][2]."'
,    '".$_POST['menu'][3]."'
);
";
?>

qphqp

i create a table like this and change table as url above code

CREATE TABLE url (
id smallint(6) NOT NULL auto_increment,
url1 varchar(50) NOT NULL default '',
url2 varchar(50) NOT NULL default '',
url3 varchar(50) NOT NULL default '',
url4 varchar(50) NOT NULL default '',
PRIMARY KEY (id)
) TYPE=MyISAM;

but i couldn't get values still.

paulnaj

Your table looks completely straighforward.

I would do a check on the variables being to save.php.

Put the following at the top of save.php (before any other code)...

<?php
echo '<pre>';
var_dump($_REQUEST);
echo '</pre>';
exit;
?>

... and paste what it returns back here on the forum.

qphqp

i got the following

array(3) {
  ["menu"]=>
  array(4) {
    [0]=>
    string(14) "www.ttt.com"
    [1]=>
    string(15) "www.zzz.com"
    [2]=>
    string(13) "www.yyy.com"
    [3]=>
    string(17) "www.xxx.com"
  }
  ["B1"]=>
  string(6) "Send"
  ["username"]=>
  string(5) "sedfe"
}

paulnaj

Ok, that looks good ... the urls we want are in the $REQUEST array (I normally use $REQUEST which contains both $GET and $POST variables).

So next, try the SQL statement again (with the tablename changed and using $REQUEST rather than $POST).

That is, take out the test code above and replace it with ...

<?php
$sql = "
INSERT INTO `url` (`id`, `url1`, `url2`, `url3`, `url4`)
VALUES (
    NULL
,    '".$_REQUEST['menu'][0]."'
,    '".$_REQUEST['menu'][1]."'
,    '".$_REQUEST['menu'][2]."'
,    '".$_REQUEST['menu'][3]."'
);
";
echo '<pre>'.$sql.'</pre>';
exit;
?>

... which is just going to print the INSERT SQL to the browser. Check to see if it's valid (paste it back here if you want).

By the way, are you using phpMyAdmin (or some other app) to build your databases?

qphqp

hi again.i am using wampserver last version so phpmyadmin;
i got this result

INSERT INTO 'url' ('id', 'url1', 'url2', 'url3', 'url4')
VALUES (
    NULL
,    'www.xxx.com'
,    'www.yyy.com'
,    'www.zzz.com'
,    'www.ttt.com'
);

paulnaj

Hi again,

Are you sure that's right?

I used backticks (`) to enclose the table-name and field-names in the SQL of my last post, but your output code seems to have single-quotes, which will not work. Did you change these?

BTW Single quotes are correct in the VALUES section of the SQL since that is a list of string values, not a list of field-names.

qphqp

yes they are backticks.in c programming sometimes single-quotes are becoming bacticks and so we were correcting them.i thouht it seems like that 🙂 now i got the following wihtout any changes.

INSERT INTO `url` (`id`, `url1`, `url2`, `url3`, `url4`)
VALUES (
    NULL
,    'www.google.com.tr'
,    'www.google.com.tr'
,    'www.google.com.tr'
,    'www.google.com.tr'
);

bradgrafelman

I'm assuming the id column is an AUTO_INCREMENT column? If so, I would recommend leaving it out of your INSERT queries entirely - let MySQL take care of that field on its own (AUTOmatically :p).

Other than that... query seems fine. You'll want to make sure you're not vulnerable to SQL injection attacks (such as by using the [man]mysql_real_escape_string/man function to sanitize all user-supplied data). Did this work for you, or are you still having problems, or ... ?

paulnaj

OK, this looks like perfectly valid SQL.

You can cut and paste this into phpMyAdmin ...

Login to phpMyAdmin
Select the database
Click the [SQL] tab
Paste in this code and click [Go]

If you get an error, something's up with your database and/or your table.

If the item get's INSERTed, there's something up with your PHP.

PS. Just saw brad's post and it's true what he says about not needing to include the 'id' field. It's perfectly valid to leave it in the SQL though, and at least this way it's explicit.

Also, very true about cleansing any variables passed to your script before using them in SQL.

qphqp

items get inserted.is the problem with my phpmyadmin.what kind of it can be?

bradgrafelman

If the items get inserted, I don't see why there's a problem..?

paulnaj

The SQL was valid ... we expected it to work and the item got inserted.

It must mean something's wrong with the PHP.

OK. Let's back up to your connection code. It says you are selecting a database called 'url'. Is that right? That's also the name of the table you created in the database.

If it is, try this 'more standard' connection code ...

<?php
$link = mysql_connect('localhost', 'root', '');
if(!$link) {
    die('Not connected: ' . mysql_error());
}

$db_selected = mysql_select_db('url', $link);
if (!$db_selected) {
    die ('Can\'t use [url] : ' . mysql_error());
}

echo 'Selected [url] database';

mysql_close();
?>

qphqp

i create my table as the following,


CREATE TABLE url (
  id smallint(6) NOT NULL auto_increment,
  url1 varchar(50) NOT NULL default '',
  url2 varchar(50) NOT NULL default '',
  url3 varchar(50) NOT NULL default '',
  url4 varchar(50) NOT NULL default '',
  PRIMARY KEY  (id)
) TYPE=MyISAM;

and i write the following to save.php

<?php
$sql = "
INSERT INTO `url` (`url1`, `url2`, `url3`, `url4`)
VALUES (
     '".$_REQUEST['menu'][0]."'
,    '".$_REQUEST['menu'][1]."'
,    '".$_REQUEST['menu'][2]."'
,    '".$_REQUEST['menu'][3]."'
);
";
echo '<pre>'.$sql.'</pre>';
exit;
?>

is there smthg wrong with them?and i don't know why we are using 2 ; at the ende of inserting values.is it wright?

paulnaj

What's your database called?

qphqp

has same name with my table

paulnaj

Try this (exactly this) ...

<?php
$link = mysql_connect('localhost', 'root', '');
if(!$link) {
    die('Not connected: ' . mysql_error());
}

$db_selected = mysql_select_db('url', $link);
if(!$db_selected) {
    die ('Can\'t use [url] : ' . mysql_error());
}

$sql = "
INSERT INTO `url` (`url1`, `url2`, `url3`, `url4`)
VALUES (
     '".mysql_real_escape_string($_REQUEST['menu'][0])."'
,    '".mysql_real_escape_string($_REQUEST['menu'][1])."'
,    '".mysql_real_escape_string($_REQUEST['menu'][2])."'
,    '".mysql_real_escape_string($_REQUEST['menu'][3])."'
);
";
$result = mysql_query($sql);
if(!$result) {
    die ('Query error : ' . mysql_error());
}

mysql_close();
?>

qphqp

great thank you!
where should i begin again to coding with php brcause ihave alot of mistakes..i am new but i have c , c++ knowledge.