How to display mysql records as preselected checkboxes? - php

Question

I have a table column called post_tags within a table called posts where assigned tags are stored separated by the # symbol. I also have a table called tags where all tag names are stored. I would like to design my database in a more normalized way but for the purpose I am trying to achieve this is the easiest option.
Anyway, I want to display on the screen all the entries from the tags table as checkboxes, so I do:
$query = mysql_query("SELECT * FROM tags ORDER BY name");
while ($row = mysql_fetch_assoc($query)) {
$tag = $row['name'];
echo "<input type='checkbox' name='tags[]' value='$tag' />\n";
}
Next I want to have the tags that are assigned to a particular post be preselected. For example, if I have a post with the following in it's post_tags column:
party#beaches#dolphins#
I want the "party", "beaches" and "dolphin" checkboxes to be checked by default (while the checkboxes for the other options are unchecked). How can this be done?

Answer 1

try the two results and the in_array() function.
<?php
$tags = mysql_query("SELECT * FROM tags ORDER BY name");
$post_tags = "party#beaches#dolphins#";
$arr_tags = explode("#", $post_tags);
while ($row = mysql_fetch_assoc($query)) {
$check = in_array($arr_tags, $row['name'])? 'checked="checked"' : "";
echo '<input type="checkbox" name="tags[]" value="'.$row['name'].'" '.$check.' />';
echo "\n";
}
?>
UPDATE
Because of Jeff question on performance, I looked for faster solutions and using isset() is faster so this would do a faster lookup of the values. the array_flip() is 3 time less taxing than in_array():
<?php
$tags = mysql_query("SELECT * FROM tags ORDER BY name");
$post_tags = "party#beaches#dolphins#";
$arr_tags = array_flip(explode("#", $post_tags));
while ($row = mysql_fetch_assoc($query)) {
$check = isset($arr_tags[$row['name']])? 'checked="checked"' : "";
echo '<input type="checkbox" name="tags[]" value="'.$row['name'].'" '.$check.' />';
echo "\n";
}
?>

Answer 2

The first thing to do is see if there is any existing data. So run that query and put the result of that table cell into lets say $checkedstring if not then put your default string in.
<?php
$checkedstring = "party#beaches#dolphins#";
//Pull from DB if exsists and set $checkedstring to that value
///
$checkeditems = explode ( "#" , $checkedstring);
$checked = array();
foreach($checkeditems as $item)
{
$checked[$item]=true;
}
$query = mysql_query("SELECT * FROM tags ORDER BY name");
while ($row = mysql_fetch_assoc($query))
{
$tag = $row['name'];
$checkedstatus = '';
if($checked[$tag])
{
$checkedstatus = "checked='checked'";
}
echo "<input type='checkbox' name='tags[]' value='$tag' $checkedstatus />\n";
}
?>