Change checkbox to checked with mysql and php

Question

my last question was no quite good, so im sorry to "open" new one.

Importante data:

Table1

id:1
id:2
id:3

Table2

id:1
id:3

Query1 (id Results):

1,2,3

Query2 (id Results):

1,3

So, I have a checkbox that need to be changed to "checked" if everytime that I fetch these 2 tables (with proper query ofc) and some id from table2 exists on table1.

Im doing like this:

<?php
while($row = mysql_fetch_array($Query1))
{
while($row2 = mysql_fetch_array($Query2))
{
$checkit = '';
if($row['id'] == $row2['id'] )
$checkit = 'checked="checked"';
}
?>
<input type="checkbox" name = "checkbox-1" class="checkbox" value ="" <?php echo $checkit ?> />
<?php
}
}
?>

Problem: I just can have "checked" for the first result, when 1=1.

That While fetch only once. I need that the result should be like this (id): 11-13 | 21-23 |31-33 and checked checkbox should appear only when 11 and 33

I hope that you understand this..

possible duplicate of how to compare two tables in mysql and php? — Marc B
– Marc B, Commented Aug 2, 2012 at 16:40
my last question was no quite good, so im sorry to "open" new one. — user1148875
– user1148875, Commented Aug 2, 2012 at 16:43
It's best to edit your previous question rather than opening a new one, that way nobody thinks you're trying to spam the site with the same one. — jprofitt
– jprofitt, Commented Aug 2, 2012 at 16:44

Abid Hussain · Accepted Answer · 2012-08-02 16:52:29Z

1

you have use

$checkit = 'checked="yes"';

should be

 $checkit = 'checked="checked"';

try this:--

<?php

    $arr_one = '';
    while($row = mysql_fetch_array($Query1))
    {

        $arr_one[]= $row['id'];

    }


    while($row2 = mysql_fetch_array($Query2))
    {
            $checkit = '';

            $id = $row2['id'];
            if (in_array($id,$arr_one))
            {
                $checkit = 'checked="checked"';
            }



    ?>
            <input type="checkbox" name= "checkbox-1" class="checkbox" value ="" <?php echo $checkit ?> />
    <?php
        }

    ?>

edited Aug 2, 2012 at 16:52

answered Aug 2, 2012 at 16:40

Abid Hussain

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3 Comments

user1148875 Over a year ago

ok but there is not the main problem. That While fetch only once. I need that the result should be like this (id): 11-13 | 21-23 | 31-33

Abid Hussain Over a year ago

I have update my answer. Take a look and let me know if not working.

user1148875 Over a year ago

forget my last question in the "comments" , it was working proper, thanks again!

Eugene H · Accepted Answer · 2012-08-02 16:53:44Z

1

You will need to reset your internal result pointer after the internal while block.

    while($row = mysql_fetch_array($Query1))
    {
        while($row2 = mysql_fetch_array($Query2))
        {
              // Your code
        }
        mysql_data_seek($Query2,0);
    }

answered Aug 2, 2012 at 16:53

Eugene H

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Comments

Justin Swanhart · Accepted Answer · 2012-08-02 17:05:29Z

1

I would do:

$t2data = array();
while($row2 = mysql_fetch_assoc($Query2)) {
  $t2data[] = $row2[0];
}

while($row1 = mysql_fetch_assoc($Query1)) {
  if(in_array($row1[0], $t2data)) { 
    $checked = 'checked';
  }
}

That being said, if both of the tables are in the same database, this is best solved with an OUTER JOIN query:

select t1.id, (t2.t1_id IS NOT NULL) as checked
  from t1 
  left join t2 on (t1.id = t2.t1_id)

Then loop over the results and check only the boxes where checked=true from the query.

edited Aug 2, 2012 at 17:05

answered Aug 2, 2012 at 17:00

Justin Swanhart

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