store the value to array, and select array intersected

Question

1.I would like to store the array value to this three variable, and take the intersected value of it, but it did not work!

$report =array();
$report1 =array();
$report2 =array();
while ($row=mysql_fetch_array($r_query)){
   echo 'Primary first State:' .$row['user_id'].'<br>';
    $report[]=array(
         'id'=>$row['user_id']);
}

while ($rowa=mysql_fetch_array($r_querya)){
   echo 'Primary second State:  ' .$rowa['user_id'].'<br>'; 
 $report1[]=array(
         'id'=>$rowa['user_id']);
}
while ($rowb=mysql_fetch_array($r_queryb)){
   echo 'Primary third State:  ' .$rowb['user_id'].'<br>';  
 $report2[]=array(
         'id'=>$rowb['user_id']);
}
echo var_dump($report);
   $result = array_intersect($report,$report1,$report2);
   echo 'intersect State:  ' .$result;
}
?>

**WARNING**: If you're just learning PHP, please, do not learn the obsolete `mysql_query` interface. It's awful and is being removed in future versions of PHP. A modern replacement like [PDO is not hard to learn](http://net.tutsplus.com/tutorials/php/why-you-should-be-using-phps-pdo-for-database-access/). A guide like [PHP The Right Way](http://www.phptherightway.com/) can help explain best practices. Always be absolutely **sure** your user parameters are [properly escaped](http://bobby-tables.com/php) or you will have severe [SQL injection bugs](http://bobby-tables.com/). — tadman, Jun 24 '15 at 17:21
"Did not work" is not a diagnostic we can help you with. Do you have any errors to report? What query are you running? Are you even checking for errors when making SQL calls? — tadman, Jun 24 '15 at 17:22
echo var_dump($report); The 3 array store the value like this array(2) { [0]=> string(1) “5” [1]=> string(1) “7” } but the intersect only output 'Array'! — Bing Chen, Jun 24 '15 at 17:37

saulnier · Answer 1 · 2015-06-24T18:00:59.757

0

The result of an array_intersect is an array itself, thus replace the last line of code with something like:

echo 'intersect State:  ' . PHP_EOL;
var_dump($result);

In addition:

echo var_dump($report);

doesn't make much sense, use only the latter (var_dump) for arrays... .

edited Jun 24 '15 at 18:00

answered Jun 24 '15 at 17:53

saulnier

83
1
4

Yes, it output the intersect value! thank you so much. In addition, can I directly use this variable to here //Country output $sqli = "SELECT meta_value FROM search WHERE user_id LIKE '%".$result."%' and meta_key='pmpro_bcountry'"; $r_queryi = mysql_query($sqli); – Bing Chen Jun 24 '15 at 18:32
//Country output $sqli = "SELECT meta_value FROM search WHERE user_id LIKE '%".$result."%' and meta_key='pmpro_bcountry'"; $r_queryi = mysql_query($sqli); while($a=mysql_fetch_array($r_queryi)){ echo '
County: '.$a[meta_value]; } – Bing Chen Jun 24 '15 at 18:51
If your result returns an array of user_id -s you can accomplish it by: `$search_list = '(' . explode(',',$result) . ')'` and placing that in your sql like: `... WHERE user_id IN " . $search_list`; – saulnier Jun 24 '15 at 19:00
The result is like this: @saulnier array(1) { [0]=> array(2) { [0]=> string(1) “5” [1]=> string(1) “7” } } – Bing Chen Jun 24 '15 at 19:04
This is the error warning: explode() expects parameter 2 to be string, array given in – Bing Chen Jun 24 '15 at 19:12
Sorry, I meant `implode` – saulnier Jun 24 '15 at 19:32
If this helped you, please indicate this as the answer. Thank you. – saulnier Jun 24 '15 at 20:03
yes, I fix it, by changing the code: $result = array_values(array_intersect($report1,$report2,$report)); – Bing Chen Jun 24 '15 at 21:47

store the value to array, and select array intersected

1 Answers1