[SOLVED] Difficult Arrays and Check Boxes

[phpretard]

My objective is to get the comma seperated values from an individual row and display them as checkboxes.

 

$all_cats[] = value1, value2, value3, value4, value5, value6 <- all should be displayed as checkboxes

 

$type[]= value2, value4,  value6 <-- these should be "checked"

 

<?php
connect();

$get_current_categories = mysql_query("select * from bugs2 where id='".$_GET['id']."' ") or die(mysql_error());
while ($cat_check = mysql_fetch_assoc($get_current_categories))
{
$type[]=$cat_check['type'];	
}

$imp=implode(", ", $type);	
$exp=explode(", ", $imp);

$get_all_categories = mysql_query("select * from types order by name ") or die(mysql_error());
while ($all = mysql_fetch_assoc($get_all_categories))
{
$all_cats[] = $all['name'];
}


foreach($all_cats as $category)
{
	if(in_array($exp, $all_cats)){ // PROBLEM IS HERE I BELIEVE
		$selected = "<input type=\"checkbox\" name=\"categories[]\" value=\"$category\" checked />$category";
	}else{
		$selected = "<input type=\"checkbox\" name=\"categories[]\" value=\"$category\" />$category";
	}

echo $selected;
}		


free($get_current_categories);	
free($get_all_categories);	

?>

 

The problem is all the chackboxes show up but none are checked.

I think I have pinpointed where the problem is but am unable to fix it.

 

Thank you

[xangelo]

You've flipped the in_array arguments.

 

if(in_array($exp, $all_cats)

 

Should actually be

 

if(in_array($all_cats, $exp)

 

The first value in in_array is what you're looking for. The second is the array you're looking for it in.

 

http://us2.php.net/manual/en/function.in-array.php

[phpretard]

It still didn't work for some reason...

 

Can you see any other issues?

[Psycho]

You are making this way more difficult than it needs to be. Simply do ONE query to get all the optins and add a parameter using JOIN to determine which ones should be checked. Give me a moment and I'll post some code based upon what you currently have.

 

However, based on what you have the 'type' field in the 'bugs2' table would need to match up with the 'name' field in the 'types' table? Is that correct?

[kickstart]

Hi

 

I agree with the above. Just been playing:-

 

<?php
connect();

$get_all_categories = mysql_query("select a.name, b.type from types a LEFT OUTER JOIN bugs2 b ON a.name = b.type AND id='".$_GET['id']."' order by a.name ") or die(mysql_error());
while ($all = mysql_fetch_assoc($get_all_categories))
{
echo "<input type=\"checkbox\" name=\"categories[]\" value='".$all['name']."' ".((is_null($all['type'])) ? '' : 'checked' )." />".$all['name'];
}

?>

 

However you should check that $_GET['id'] is legit before using it in SQL.

 

All the best

 

Keith

 

All the best

 

Keith

[phpretard]

I "hardcoded" the id to 20(known to have the right info) and I get:

 

Column 'id' in on clause is ambiguous

 

What does that mean?

 

I am researching as you (will hopefully) respond.

 

THANK YOU!!!

[kickstart]

Hi

 

Means that more than one of the tables has a column called id. So it doesn't know to check types.id or bugs2.id

 

Qualify the id field for this:-

 

<?php
connect();

$get_all_categories = mysql_query("select a.name, b.type from types a LEFT OUTER JOIN bugs2 b ON a.name = b.type AND b.id='".$_GET['id']."' order by a.name ") or die(mysql_error());
while ($all = mysql_fetch_assoc($get_all_categories))
{
echo "<input type=\"checkbox\" name=\"categories[]\" value='".$all['name']."' ".((is_null($all['type'])) ? '' : 'checked' )." />".$all['name'];
}

?>

 

All the best

 

Keith

[Psycho]

This should be all that you need. May need to check that field names are correct for your environment. It basically does a select all names on the 'types' table and then appends a new field to each record ('checked' with a value of 1 or 0) based upon whether there is a match in the 'bugs2' table with the type value matching the name.

 

$query = "SELECT name,
            IF (name IN (SELECT type FROM bugs2 WHERE id='{$_GET['id']}'),1,0) as checked
          FROM types
          ORDER BY name"
$result = mysql_query($query) or die(mysql_error());

$checkboxes = '';
while ($checkData = mysql_fetch_assoc($result))
{
    $checked = ($checkData['checked']==1) ? ' checked="checked"' : '';
   $checkboxes .= "<input type="checkbox" name="categories[]" value="{$checkData['name']}"{$checked} />$checkData['name']<br /> ";
}

echo $checkboxes;

 

EDIT: kickstart's method should work too. Just different ways to accomplish the same thing. I suspect his may be more efficient, but don't know for sure.

[phpretard]

Is this code to replace part or all of the existing?

 

I ask because it still gives me all blank checkboxes...although it does display all the checkboxes

 

The (comma seperated) value of b.type='value2, value4, value6'

 

Should I not comma seperate?

 

That is some great code by the way!

 

It all makes sense to me except for:

 

--- ".((is_null($all['type'])) ? '' : 'checked' )." ---

 

I don't quite understand why the question mark and the colon are there.

[kickstart]

Hi

 

The method is very similar to mine, just using a subselect while I have used a join.

 

It was to replace you whole fragment of code.

 

The bit  ".((is_null($all['type'])) ? '' : 'checked' )." is doing an IF within a statement. It checks if the returned type feield is null with is_null($all['type']) , if that is true it returns the bit after the ? (ie, nothing in this case), if false it returns the bit after the : ('checked').

 

Where is he comma seperated list of b.type coming from?

 

All the best

 

Keith

[phpretard]

Comma seperation comes from:

 

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

$categories=implode(", ", $_POST['categories']);

connect();
$insert=mysql_unbuffered_query("update bugs2 set type = '$categories' where id='".intval($_POST['id'])."' ") or die(mysql_error());

}

 

 

So when I update it changes the values in the DB  (with the simplified code Hooray!)

 

it's the display that's killing me...it still won't show "checked" boxes

 

[kickstart]

Hi

 

Arrggh. That wasn't in the original post.

 

What is it that you want from that table? What are categories? Why are they stored in one field?

 

My guess is that you have bugs which have various catagories. If so what you should really have is a table of bugs and another table list all the categories for each bug.

 

All the best

 

Keith

[phpretard]

I apologize for that...

 

I am trying to fix someone elses mess and it seemed the easiest way to resolve was to comma seperate in the (field "type") in the table "bugs"), then reference the table I added called (categories) that has six entries (which are the categories).

 

[attachment deleted by admin]

[kickstart]

Hi

 

If you can move the types off to a seperate table then that would be far easier, as you can then do most of the work in one SQL call.

 

While there are ways to extract the comma seperated field, they are not nice to read, would be a nightmare to maintain and I would perform badly. As well as giving you major problems if someone introduces a type with a comma in it.

 

To continue with what you already have:-

 

<?php
connect();

$TypeArray = array();
$get_current_categories = mysql_query("select type from bugs2 where id='".$_GET['id']."' ") or die(mysql_error());
while ($cat_check = mysql_fetch_assoc($get_current_categories))
{
$types = explode(',',$cat_check['type']);
foreach($types as $type)
{
	$type = trim($type);
	$TypeArray[$type] = $type;
}
}


$get_all_categories = mysql_query("select name from types order by name ") or die(mysql_error());
while ($all = mysql_fetch_assoc($get_all_categories))
{
echo "<input type='checkbox' name='categories[]' value='".$all['name']."' ".((isset($TypeArray[$all['name']])) ? "checked='checked'" : '' )." />".$all['name'];
}

free($get_current_categories);	
free($get_all_categories);	

?>

 

All the best

 

Keith

[phpretard]

You are right...

 

I was trying to save a little time but look how much I've spent (yours and mine) with band-aids

 

Thank you for your help!!!

 

I will just rebuild it the right way.