Php mysql query and foreach does not work inside function; but outside function the same code works

Question

Here php function

(Update after reading comments etc.)

Yes, $db is defined outside and above function

$db = new PDO("mysql:host={$dbhost};dbname={$dbname};charset=utf8", $dbuser, $dbpass//, array(PDO::ATTR_EMULATE_PREPARES => false, PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION));

function vatRates () {
   $something = "test";
   echo $something;

try {
   $query_select_date = "SELECT CurrencyRate FROM 2013Currencies WHERE DateOfCurrencyRate = '2001-03-23'";

   $sql_select_date = $db->prepare($query_select_date);
   $sql_select_date->execute();
   $data_select_date = $sql_select_date->fetchAll(PDO::FETCH_ASSOC);
   }
 catch (PDOException $e){
 echo "<br>DataBase Error: " .$e->getMessage();
 }
 catch (Exception $e) {
 echo "General Error: ".$e->getMessage() .'<br>';
 }

   foreach ($data_select_date as $data) {
       echo($data[CurrencyRate]);
   }


}

And then call the function with vatRates();

In output get only word: test from $something. And php code that is after vatRates(); does not execute.

But if delete function vatRates () { and closing } then mysql query and foreach works.

Why mysql query and foreach does not work inside function (what need to correct)?

Where do you get $db? Why don't you have error reporting on? — Your Common Sense
– Your Common Sense, Commented Aug 2, 2013 at 12:18
Are you missing a global $db at the start of the function, by any chance? When using functions, you have to specify what variables will require global scope. — NoGall
– NoGall, Commented Aug 2, 2013 at 12:18
$db is not defined inside the scope of the function: pass it as an argument — Mark Baker
– Mark Baker, Commented Aug 2, 2013 at 12:18
@LokiSinclair global should not be the first recommendation to resolve function scope problems — Mark Baker
– Mark Baker, Commented Aug 2, 2013 at 12:19
aside from the obvious missing $db parameter, is CurrencyRate defined? You probably don't want to fill up your log with 'undefined CurrencyRate using "CurrencyRate" instead' — Kris
– Kris, Commented Aug 2, 2013 at 12:24

JohnnyFaldo · Accepted Answer · 2013-08-02 12:20:13Z

2

function vatRates ($db) {
   $something = "test";
   echo $something;

   $query_select_date = "SELECT CurrencyRate FROM 2013Currencies WHERE DateOfCurrencyRate = '2001-03-23'";

   $sql_select_date = $db->prepare($query_select_date);
   $sql_select_date->execute();
   $data_select_date = $sql_select_date->fetchAll(PDO::FETCH_ASSOC);

   foreach ($data_select_date as $data) {
       echo($data[CurrencyRate]);
   }

}

call with vatRates($db)

Reason: $db is currently outside the visibility of your function, this passes it to the function.

answered Aug 2, 2013 at 12:20

JohnnyFaldo

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Comments

Mohammad Masoudian · Accepted Answer · 2013-08-02 12:19:27Z

2

function vatRates () {
   global $db;
   $something = "test";
   echo $something;

   $query_select_date = "SELECT CurrencyRate FROM 2013Currencies WHERE DateOfCurrencyRate = '2001-03-23'";

   $sql_select_date = $db->prepare($query_select_date);
   $sql_select_date->execute();
   $data_select_date = $sql_select_date->fetchAll(PDO::FETCH_ASSOC);

   foreach ($data_select_date as $data) {
       echo($data[CurrencyRate]);
   }

}

answered Aug 2, 2013 at 12:19

Mohammad Masoudian

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1 Comment

AmazingDreams Over a year ago

Recommend passing $db as a parameter.

Alexandre Ouicher · Accepted Answer · 2013-08-02 12:20:35Z

2

Function must be return something, a string, a array, a bolean, etc..

Better is (with arg $db passed):

function vatRates ($db) {
$CurrencyRate = "";

$query_select_date = "SELECT CurrencyRate FROM 2013Currencies WHERE DateOfCurrencyRate = '2001-03-23'";

$sql_select_date = $db->prepare($query_select_date);
$sql_select_date->execute();
$data_select_date = $sql_select_date->fetchAll(PDO::FETCH_ASSOC);

foreach ($data_select_date as $data) {

$CurrencyRate .= $data[CurrencyRate];
}
return $CurrencyRate;
}

To use it:

echo vatRates($db);

answered Aug 2, 2013 at 12:20

Alexandre Ouicher

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2 Comments

AmazingDreams Over a year ago

This is indeed 'the better way' to do it, it is no 'must do' though.

user2465936 Over a year ago

Regarding return here php.net/manual/en/function.return.php read that ends execution of current function. What if instead return $CurrencyRate; also need return $Unit; and other variables? return $CurrencyRate. $Units. $Etc;

som · Accepted Answer · 2013-08-02 12:19:09Z

1

function vatRates () {
   $something = "test";
   echo $something;
   global $db;

   /* Your sql code and foreach goes here*/

}

or

function vatRates ($db) {
       $something = "test";
       echo $something;

       /* Your sql code and foreach goes here*/

 }

answered Aug 2, 2013 at 12:19

som

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1 Comment

AmazingDreams Over a year ago

Recommend passing $db as a parameter.

jakecraige · Accepted Answer · 2013-08-02 12:19:15Z

0

Could be an issue with you not accessing the values in the array correctly. Try changing the foreach to this

foreach ($data_select_date as $data) { 
  echo($data['CurrencyRate']);
}

answered Aug 2, 2013 at 12:19

jakecraige

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