Argument 1 passed to myFunction() must be an instance of string, string given, called in

Question

i have this function:

function myFunction(string $name) {
    $db = mysql_connect("localhost", "root", "");
    mysql_select_db(...); 

    $insertplayer="INSERT INTO `...`(....)
    VALUES (......')";
    if (!mysql_query($insertplayer,$db))
      {
      die('Error: ' . mysql_error());
      }
    $id = mysql_insert_id();
    echo 'done for player N°'.$id;
    mysql_close($db);
}

and the form i use:

<form action="insertplayer.php" method="post">
    <input type="text" name="nameplayer" />
    <input type="submit" value="ok" />
</form>

But when i do this, i have this error:

Catchable fatal error: Argument 1 passed to myFunction() must be an instance of string, string given, called in C:.... on line 23 and defined in C:...

I have this error with the int problem. How can i resolved it ?

you don't need to use dataType string in php in your function — Awlad Liton
– Awlad Liton, Commented Mar 13, 2014 at 16:18
Type hinting is not for primative types, php is looking for a string class that does not exist. Remove that type hint: myFunction($name) { — Steve
– Steve, Commented Mar 13, 2014 at 16:18
Can you show us the code where you call the function you defined ? — Lapixel
– Lapixel, Commented Mar 13, 2014 at 16:19
possible duplicate of Really PHP? "Argument 1 passed to my_function() must be an instance of string, string given" — Jason OOO
– Jason OOO, Commented Mar 13, 2014 at 16:19
Thanks, i will just need remove string in myfunction(string $....) — user3414480
– user3414480, Commented Mar 13, 2014 at 17:40

Deep in the Code · Accepted Answer · 2015-02-25 17:09:08Z

4

Try removing the primitive data type from the function declaration. PHP does not require type hints for primitive data types.

function myFunction($name) {
    $db = mysql_connect("localhost", "root", "");
    mysql_select_db(...); 

    $insertplayer="INSERT INTO `...`(....)
    VALUES (......')";
    if (!mysql_query($insertplayer,$db))
      {
      die('Error: ' . mysql_error());
      }
    $id = mysql_insert_id();
    echo 'done for player N°'.$id;
    mysql_close($db);
}

answered Feb 25, 2015 at 17:09

Deep in the Code

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Comments

Francesco Casula · Accepted Answer · 2015-11-18 09:55:05Z

0

If you don't want to do a block of code to check type at the beginning of each function you are writing looking like this:

function myFunction($myScalar) {
    if (!is_string($myScalar)) {
        throw new Exception(...);
    }

    ...
}

You can register an errorhandler comparing the given type with your hint and if for example string==string you can ignore the error.

See User Notes on the manual page for some examples

http://php.net/manual/en/functions.arguments.php#functions.arguments.type-declaration

edited Nov 18, 2015 at 9:55

Francesco Casula

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answered Feb 25, 2015 at 19:15

DerOetzi

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