sending date variable to another page doesn't work always

[RomeoX]

hi everybody.

I have a question that I have two pages in php, the first one for editing the date in a table. when I enter a date, doesn't want to update the record that I want so I found that only one record out of 10 is updating and the rest doesn't want to update. I feel that there is something wrong.

my first page code looks like that in brief

<td class="style4"><?php echo $row['libraryid']; ?></td>
    <td class="style4"><?php echo $row['dateborrow']; ?></td>
    <td class="style4"><input name="returned" type="text" size="15" id="returned" value="<?php echo $row['datereturned']; ?>" /></td>
    <td class="style4">SR <?php echo $row['amountoffine']; ?></td>

    <td><input name="c[]" type="checkbox" id="c[]" value="<?php echo $row['borrowid']; ?>" /></td>
  </tr>
  <?php
  }
  ?>
</table>
<center>
<p>
  <input name="checkfine" type="submit" value="Check For Fine" />

see this line <input name="returned" type="text" ...........

that I pass the date value through text field to another page.

here is the code of the another page:

<?php
include 'functions.php';
$submit = $_POST['checkfine'];
$returned = date($_POST['returned']);   // <----Is the syntact correct here
$datereturned = $_POST['datereturned'];

if (isset($submit))
		  {
$id = $_POST["c"]; 
foreach($id as $v)
{

$queryreg = mysql_query("UPDATE bookoutonloan SET datereturned = '$returned' where borrowid = '$v' ");

}
//die("The Record Date Has Been Updated");
		  }
?>

This code doesn't help me always, I don't know where is the problem. Please help me in identifying the problem.

[Ashley_Sheridan]

Look at the HTML output in your browser, I suspect that it's either a malformed string that is breaking the input element, or a malformed date that isn't inserting into the database.

Also, turn errors on so you can see exactly what (if any) errors are being generated.

[bradgrafelman]

You're using the 'returned' form element in the [man]date/man command, which is used to format the current date and time (if no timestamp is given as a second parameter).

If you just want the date submitted, then don't use any PHP function at all - just use the string sent in the form.

[RomeoX]

Thanks to all,

What you meant by malformed? Another thing I have the debug error on and there is no error.

Actually I tried with out date function but doesn't want to insert the date

I already tried this:
$returned = $_POST['returned'];

================

I did this way to use a date directly and I found that it's working but it's not the way that I'm seeking

$returned = $_POST['returned'];

if (isset($submit))
		  {
$id = $_POST["c"]; 
foreach($id as $v)
{

$queryreg = mysql_query("UPDATE bookoutonloan SET datereturned = '2010-03-04' where borrowid = '$v' ");

So I think there is something I have to use so the value of the date will be tranmitted to another page but I don't know what exactly

[bradgrafelman]

What column type is 'datereturned' and what does $_POST['returned'] look like if you echo it out to the screen?

[RomeoX]

thanks again for helping me.

Actually I tried your suggestion and I'm happy with it that I try to echo in the page and see the result and I found that nothing print from the first record. I found something that if I enter data in the last record it print and it's fine. So after I tested many things I come up with:
-When I // the while and // the } for the while and I tested and then I can echo because there will be only one TextFiled so it ofcourse it will print.

So I think the problem in CheckBox that doesn't specify which record has to be updated. So what should I do?

[bradgrafelman]

RomeoX;10944900 wrote:

So I think the problem in CheckBox that doesn't specify which record has to be updated. So what should I do?

Confirm your suspicions by echo'ing out the SQL queries so that you can visually inspect them.

[RomeoX]

the sql query in the second page doesn't show anything. Is that what you are asking me about or something else

[bradgrafelman]

I mean echo out the SQL query string so you can visually inspect it.

[RomeoX]

Could you please write the the the whole echo that you are mentioning?

Actually I think that the Checkbox is not taking the exact value for specific record.

Here it's:
$id = $_POST["c"];
foreach($id as $v)

So I think this checkbox is not enough because can't take the value from the previous page, even in the previous page there is CheckBox with the "Name:c[]" but doesn't work properly.

[Ashley_Sheridan]

Instead of doing

mysql_query("UPDATE bookoutonloan SET datereturned = '$returned' where borrowid = '$v' "); 

Do:

$query = "UPDATE bookoutonloan SET datereturned = '$returned' where borrowid = '$v' ";
echo $query;
mysql_query($query); 

Assigning your query string to a variable first is the best practice I've found, as when you run into a situation where your query doesn't seem to be doing what you think it should be, it's very easy to output it with an echo statement.

[RomeoX]

Thanks Ashley for explaining to me.

Okay now after executing the ECHO QUERY I got this result in the output

UPDATE bookoutonloan SET datereturned = '' where borrowid = '15' 

So the datereturned is empty.

=========================
Actually I was thinking that the TextFiled has to have array [] like what I did in CheckBox so I changed something like this or what I have to sign it exactly here?.

<input name="returneed[]" type="text" size="15" id="returneed[]" />

And I change the Foreach with For code like this in the second page:

<?php
include 'functions.php';
$submit = $_POST['checkfine'];
//$returned = $_POST["returneed"];

if (isset($submit))
	  {
	for($i=0;$i<count($_POST['c']);$i++)

 {
    for($b=0; $b<count($_POST['returneed']); $b++)

                {


$query = ("UPDATE bookoutonloan SET datereturned = '".$_POST['returneed'][$b]."' where borrowid = '".$_POST['c'][$i]."' ") ;

echo $query;
$final = mysql_query($query) or die(mysql_error()); 


                }

            }

}
?>

And this is the result that I get in the output by using Echo:

UPDATE bookoutonloan SET datereturned = '2010-04-22' where borrowid = '15' UPDATE bookoutonloan SET datereturned = '' where borrowid = '15' 

The result that I got in the output by using Echo I got two UPDATE the first one I got the date that I entered and the second UPDATE showing nothing, and when I went to check in my database I found it 0000-00-00. So still I'm having problem. I hope if you could solve it please. Thanks for all for cooperation.

[Ashley_Sheridan]

Well, you now have two queries running to update the date. The first one is correct and will set it, but the second one is overriding what you've just done and setting the date field to an empty string, which is interpreted by MySQL as '0000-00-00'

Do some sort of check before you run the update query to ensure that the date you think should be there is actually there. A regex check in the form of

/^\d{4}\-\d{2}\-\d{2}$/

should do it.

[RomeoX]

Thanks Ashley

Excuse me, what should I do with this code: /^{\d{4}-\d{2}-\d{2}$/}
I mean where should I paste it, hope if you could give me the script to added urs as well. I think this will do some formatting for date input. But how do I get rid of the second update query that prevent my date to be added.

===============================
Something else I want to mention that now I have four records retrieving from DB
when I enter the date for the first 3 I get the result as I said before 2 updates in the output like this way will be for borrowid15,16,17:

UPDATE bookoutonloan SET datereturned = '2010-04-22' where borrowid = '15' UPDATE bookoutonloan SET datereturned = '' where borrowid = '15'

but for borrowid 14 the last record the output becomes opposite like this:

UPDATE bookoutonloan SET datereturned = '' where borrowid = '14' UPDATE bookoutonloan SET datereturned = '2010-05-05' where borrowid = '14' 

I don't know why it happens like that.

[Ashley_Sheridan]

It's a regex, so you would use it with one of the PHP regex functions, such as preg_match().

So, to test if the string sent in the get data is a valid date in the format you expect, you could do:

$date = (preg_match('/^\d{4}\-\d{2}\-\d{2}$/', $_GET['date']))?$_GET['date']:'0000-00-00';

It's untested, but basically it checks to see if the date is in the right format. If it is, it sets $date to the given value, if not, it sets $date to a default '0000-00-00'.

The regex itself is fairly simple, \d means a digit, and the numbers in braces indicate how many of that character you want to match. The hyphens are escaped with a \ because they have special meaning in regular expressions.

I tend to validate things in this way if I know the exact format they should take, like dates, postcodes, etc. Always get into the habit of validating your data, it's probably the most important thing you could ever do in a web app.

[RomeoX]

excuse me Ashley, Could you please read again my previous post I edited now. It has extra information I added. Thanks

[RomeoX]

Anyone got any solution for my code, I really need help today. Please Please 🙁🙁🙁🙁

[RomeoX]

still looking for solution.