Multiple update is not working

sal

If anyone could tell me why the following is not updating please

form1.php

<?
$sql = mysql_query("select category from catlisting") or die("Unable to process queries at this time: " . mysql_error());
while ($i = mysql_fetch_array($sql))
{
$category = $i["category"];
$result .= "<OPTION value=\"$category\">$category</OPTION>";
}
?>
<form action=form2.php?l=<?php echo $l; ?>&username=<?php echo $username; ?>&password=<?php echo $password; method=post enctype="post">
<?
$sql = mysql_query("select * from categories WHERE username = '$username'") or die("Unable to process queries at this time: " . mysql_error());
while ($i = mysql_fetch_array($sql)) {

?> <select class=select name="category[]">
<option value="<? echo "$i[category]"; ?>" ><? echo "$i[category]"; ?> </option>
<? echo "$result"; ?>
</SELECT>
<?
}
?>

form2.php

foreach ($_POST["category"] as $newcat) {

if ($newcat !="") {
$sql = "UPDATE categories SET category='$newcat' WHERE username = '$username'";
}
}

Here is the result of the update query I get:

UPDATE categories SET category='111' WHERE username = 'debbie'
UPDATE categories SET category='222' WHERE username = 'debbie'

Tells me it's updating and yet, it isn't

artic

you got it as

<form action=form2.php?l=<?php echo $l; ?>&username=<?php echo $username; ?>&password=<?php echo $password; method=post enctype="post">

should be

<form action=form2.php?l=<?php echo $l; ?>&username=<?php echo $username; ?>&password=<?php echo $password;?> method=post enctype="post">

and

i dont notice where the sql database is started in second form and inserted if im reading thsi right.

cgraz

$sql = "UPDATE categories SET category='$newcat' WHERE username = '$username'";

That's just a string, you haven't actually executed it. Your database won't be updated because that's a string. execute it using mysql_query()

Cgraz

sal

Thanks. I did forget to uncomment it. Problem is that when I do, all values are replaced. Instead of the values entered I get Array!

cgraz

That's because your select is an array

drop the [] and you'll get the actual value instead of the word Array. They can only select one option anyway; what's the point of making it an array?

Cgraz

sal

<form action=form2.php?l=<?php echo $l; ?>&username=<?php echo $username; ?>&password=<?php echo $password;?> method=post enctype="post">

is what I have. Must have copied and pasted poorly

.... which is a reflexion on my programming skills!

Am still not able to update.

However, if, instead of the select tag in form 1, i use an input tag then I do get an update!

cgraz

see my last post

Cgraz

sal

That's not even true any more. I used to be able to update with just input and not select but not anymore!

I guess I need to go to bed.

cgraz

does that mean it's working now?

Cgraz

sal

Here is what I get

in form1.php

I replaced the select tag with an input tag like so:

$sql = mysql_query("select * from categories WHERE username = '$username'") or die("Unable to process queries at this time: " . mysql_error());
while ($i = mysql_fetch_array($sql)) {
?>
input name="category[]" value="<? echo "$i[category]"; ?>
<?
}
?>

In form2.php

foreach ($_POST["category"] as $newcat) {
$sql = "UPDATE categories SET category='$newcat' WHERE username = '$username'";
mysql_query($sql);
}

So.. If I enter the same values in form1.php like

111
111

The update works

likewise for

222
222

The update works

However if I change the values

111
222

Then the update does not work

cgraz

do you have a link where i could see this? your code looks wrong here

input name="category[]" value="<? echo "$i[category]"; ?>

should be

Like i mentioned earlier, you don't need the name to have the brackets [] at the end. You can also drop the foreach loop and do

$sql = "UPDATE categories SET category='$category' WHERE username = '$username'";

You can only input one value into a text field, right? No need for an array.

Cgraz

sal

If I change

And drop the foreach on the second form, then the update only works once.

If I restore the []

the second value I enter in form 1 replaces both values.

11
22

I get

22
22

or
22
11

I get
11
11

I am sorry about this nonsense by the way! I appreciate your help enormously

cgraz

I'm just confused as to why you are updating one field in a database with two values. Can you explain what the code is supposed to do? You can only enter one value into a text type field. That's why I said not to have it be an array. If you're updating one field, you're only goign to have one value.

Maybe I just need more explanation.

Cgraz

sal

I want to do this:

Debbie is a student. On her first term she took the following classes:

math
physics

Then she decided to change her courses

Math
Biology

I want to be able to update both her classes.

The table looks like

id, username, category

So from

1 Debbie math
2 Debbie physics

I want to have

1 Debbie Math
2 Debbie Biology

cgraz

Ok, tha tmakes more sense. I'm a little tired, but this hsould work. basically here's the logic behind it, with a little bit of real code

<?

select * from table where bla

?>

<form method=". . .">

<?

while($row = mysql_fetch_array($query)) {
   $i = "1"
   echo "Class " . $i . " <input type=\"text\" name=\"" . $row["id"] . "\" value=\"" . $row["category"] . "\">\n";
   $i++;
}

?>

</form>

<?

// then to process it

foreach($_POST as $key => $value) {
   $query = mysql_query("UPDATE your_table SET category='$value' WHERE id='$key'") or die(mysql_error());
}

?>

Make sure the only fields in your form are the text fields to update the categories or this won't work.

Although there may be diff users in your db, each record has a unique id. Each new class is a new record; thus having it's own id. Debbie holds id's 1 and 2, and you're only updating id's 1 and 2

Cgraz

sal

Bless you mate!

It worked, Now I have to try with a select rather than an input and I 'll be where I want to be.

Rather than write the value of the class, I want to be able to choose from a drop down list taken from a different table

Thank you again!

sal

Bless you mate!

It worked, Now I have to try with a select rather than an input and I 'll be where I want to be.

Rather than write the value of the class, I want to be able to choose from a drop down list taken from a different table

Thank you again!

sal

I did the mofications as per your suggestions:

Form 1:

<form action=form2.php?l=<?php echo $l; ?>&username=<?php echo $username; ?>&password=<?php echo $password; ?> method=post enctype="post">

$sql2 = mysql_query("select * from catlisting") or die("Unable to process queries at this time: " . mysql_error());
while($row2 = mysql_fetch_array($sql2)) {

$cat = $row2["category"];
$result2 .= "<OPTION value=\"$cat\">$cat</option>";
}

echo "<table width=90% border=1><tr>";

$sql = mysql_query("select * from categories WHERE username = '$username'") or die("Unable to process queries at this time: " . mysql_error());

hile($row = mysql_fetch_array($sql)) {

$i = "1";

echo "<td><div align=center><select name=\"" . $row["id"] . "\" value=\"" . $row["category"] . "\">";
echo "<option value=\"" . $row["id"]. "\"> " . $row["category"] . "";
echo "$result2";
$i++ ;
}

}

echo "</select></div></td></tr><tr><td colspan=4 align=center valign=middle>";
	echo "<input type=\"submit\" value=\"" . $lang[1017] . "\"></td></tr>";


echo "</table>";

?>
</form>

In form2

foreach($_POST as $key => $value) {
$query = mysql_query("UPDATE categories SET category='$value' WHERE id='$key'") or die(mysql_error());
}

This works well. HOwever, if I go back to previous page using the back button on browser, do not make any change to any of the fields and hit the submit button, all the input in the fields are replaced by the name of the field.

ie if I had: id, category
82 , Algebra
83, Chemistry

when i hit the submit button without changer the categories the result I get is

              82,82
              83,83

Is there a way to keep the old values in place?

sal

Sorry, don't bother with this one. I woked it out.

Thanks anyway