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up vote 0 down vote favorite

can anybody let me know why array_search doesnt works for me? All i want is to search value and and get corresponding key value for eg if i search wiliam i should get 4.. Its simple but aint working for me

<?php
$fqlResult[0]['uid']='1';
$fqlResult[0]['name']='Jay';
$fqlResult[1]['uid']='2';
$fqlResult[1]['name']='UserName2';
$fqlResult[2]['uid']='3';
$fqlResult[2]['name']='Frances';
$fqlResult[3]['uid']='4';
$fqlResult[3]['name']='William';



        for($i=0;$i<count($fqlResult);$i++)
        {

            $userdbname="'".$fqlResult[$i]['name']."'";
            $userdb[$userdbname]="'".$fqlResult[$i]['uid']."'"; 

        }


echo "<pre>";
print_r($userdb);
echo "</pre>";
echo array_search('4', $userdb);
?>
share|improve this question
you are adding extra quotes that are being stored inside the variable... not sure if you are doing this on purpose or not. $userdbname="'".$fqlResult[$i]['name']."'"; means that $userdbname will print as 'userName' (with the extra single quotes around it, and not just as userName). – T. Brian Jones Jul 12 '11 at 4:18
Not really related to the question, but keep in mind that if the names aren't unique, you could run into problems doing this. – Matthew Jul 12 '11 at 4:22

7 Answers

up vote 3 down vote accepted

It doesn't work because array_seach searches values and "William" is a key. To complicate things, your values and keys are wrapped in single quotes during the for loop.

You'd want to do something like this:

if ( ! empty($userdb["'William'"]) )
{
  // Echoes "'4'"
  echo $userdb["'William'"];
}

// To find user ID "'4'"
// Outputs "'William'"
echo array_search("'4'", $userdb);

If you don't want things wrapped in single quotes, you'll need to change your for loop as follows:

for($i=0;$i<count($fqlResult);$i++)
{
  $userdbname=$fqlResult[$i]['name'];
  $userdb[$userdbname]=$fqlResult[$i]['uid']; 
}

if ( ! empty($userdb["William"]) )
{
  // Outputs "4" (without the single quotes)
  echo $userdb["William"];
}

// To find user ID "4" (without the single quotes)
// Outputs "William"
echo array_search('4', $userdb);
share|improve this answer
sorry i edited my question, i wanted to type search uid=4 – Rachit Jul 12 '11 at 4:13
1  
@Rachit - That would echo "'4'". You could make it do $user_id = $userdb["'William'"]; to get the user ID. – Francois Deschenes Jul 12 '11 at 4:14
@Rachit - I updated my answer. First, your User ID is wrapped in single quotes, that's why it's not working. See my first answer. To prevent that, see my second answer. – Francois Deschenes Jul 12 '11 at 4:20
what if name has space like WIlliam Oxford or anyother special character? i used single quote to escape characters – Rachit Jul 12 '11 at 4:21
@Rachit - You don't have to do that to store data in an array. – Francois Deschenes Jul 12 '11 at 4:22
up vote 0 down vote

array_search() searches values, not keys.

If you want to check the existence of something that you use as a key in an array, you can just use isset:

if(isset($userdb['William'])) {
    echo "$userdb[William] is William's uid!";
}
share|improve this answer
sorry i edited my question, i wanted to type search uid=4 – Rachit Jul 12 '11 at 4:14
up vote 0 down vote 
for($i=0;$i<count($fqlResult);$i++)
        {

            $userdbname=$fqlResult[$i]['uid'];
            $userdb[$userdbname]=$fqlResult[$i]['name']; 

        }
share|improve this answer
I have edited my question, it still doesnt works for me – Rachit Jul 12 '11 at 4:14
it may be due to "'" you are using in loop. You need not to use them – Gaurav Jul 12 '11 at 4:16
up vote 0 down vote

Change

$userdb[$userdbname]="'".$fqlResult[$i]['uid']."'";

with this

$userdb[$i] = "{$fqlResult[$i]['name']}";
share|improve this answer
up vote 0 down vote

array_search only works with arrays of scalar data. You're trying to search an array of arrays. You can easily search the array yourself:

function search_array_col($array, $col, $val)
{
   foreach ($array as $key => $a)
   {
     if ($a[$col] == $val) return $key;
   }

   return null;
}

echo search_array_col($fqlResult, 'name', 'William') , "\n";
echo search_array_col($fqlResult, 'uid', '4') , "\n";

Edit: n/m, I misread your code. However, you could still use this to search your original array, so I'll leave the answer for reference.

share|improve this answer
up vote 0 down vote

try this:

foreach($fqlResult as $result)
{
    $name = $result["name"];
    $uid = $result["uid"];
    $userdb[$name] = $uid;
}

then you want to use array_key_exists() to find the key. array_search() only works for searching values, not keys.

$nameExists = array_key_exists("William",$userdb);
share|improve this answer
up vote 0 down vote

You can remove the quotes in the $userdbname="'".$fqlResult[$i]['name']."'"; rewrite it to

$userdbname= $fqlResult[$i]['name'];

share|improve this answer

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