Query result to array question

PHP
1

Hello forums I’m trying to put a query result to an array.

I have a table in my mysql database with the following data :

1,6,7

$q = mysql_query("SELECT bookmark FROM table WHERE id = 1 ");
$r = mysql_fetch_array($q);

$bookmark = array($r);

if (in_array(1, $r)) {
    echo 'true';
}else{
echo "false";
}

This one keeps echoing false. Any ideas? Thanks

2

Well yeah I just created another table for bookmarks and just did a left join

3

Noted thanks

4

^ This

5

What you really need to do is normalize your database :wink:

6

if (in_array(12, $w))<– may give you correct results but logically 12 is an int but from mysql it becomes string so best practice is if (in_array(“12”, $w))

7

Nevermind I got it I just need to explode the result

$w = explode(",",$r['bookmarks']); 

if (in_array(12, $w)) {
    echo "true";
}else{
echo "false";
}

<–echos false

Thanks

8

It should work if bookmark for id=1 is ‘1,6,2’

$q = mysql_query("SELECT bookmark FROM table WHERE id = 1 "); 
$r = mysql_fetch_array($q); 

$r = explode(",",$r[0]);

if (in_array("1", $r)) { 
    echo 'true'; 
}else{ 
echo "false"; 
}
9

I did this test without the query stuff and it’s working

$w = array("1","6","2");

if (in_array(6, $w)) {
    echo "true";
}else{
echo "false";
}

<– echos true. So how do I make the mysql result to display like :

$r =array(“1”,“6”,“2”);

Do I need to split it?

10

Nope I am not asking mysql to display id only bookmarks which is 1,6,7

I print_r-ed the result and it gives me this:

Array ( [0] => 1,6,2 [bookmarks] => 1,6,2 ) so it is an array

when I run the in_array() function for 1 it echoes true, but when I run in_array() for 6 and 2 it displays false. bump.

if (in_array(1, $r)) {

    echo 'true';

}else{

echo "false";

}

echos true

if (in_array(2, $r)) {

    echo 'true';

}else{

echo "false";

}

echos false… :sick:

11

An easy way to debug this would have been to simply do:

print_r ($r);

to show what was in the array.

12

You are not asking MySQL to SELECT id but you are asking to display id in the results.

Try this:



  $q = mysql_query("SELECT id, bookmark FROM table WHERE id = 1 ");

Also insert these two lines, the results are usually very informative.


  
  error_reporting(E_ALL | E_NOTICE);
  ini_set('display_errors', LOCALHOST ? 'On' : 'Off');

.