Parse error: syntax error T_VARIABLE

Question

Following is the code:

class Db{
    protected static $connection;
    public function connect(){
        if (!isset(self::$connection)) {
            $config = parse_ini_file('../config.ini');
            self::$connection = new mysqli('localhost',$config['username'],$config['password'],$config['dbname']);
        }
        if (self::$connection === false) {
            die("Error in database connection. Please contact network administrator.");
            return false;
        }
        return self::$connection;
    }
    public function quote($value){
        $connection = $this -> connect();
        return "'".$connection -> real_escape_string(string $value)."'";
    }
}

I am getting the following error in the quote function: Parse error: syntax error, unexpected '$value' (T_VARIABLE) in D:\xampp\htdocs\fizzy\submit.php on line 48

Is `$value` an array? Why are you trying to cast it as a `string` -> `string $value`? — Sean, Nov 02 '15 at 07:19
without casting, I get the following warning: Warning: mysqli::real_escape_string() expects parameter 1 to be string, array given in D:\xampp\htdocs\fizzy\submit.php on line 48 — Shreesh Katyayan, Nov 02 '15 at 07:38
Sorry, error was in function calls and not the function definition. Mark closed :) Thanks for help anyways — Shreesh Katyayan, Nov 02 '15 at 08:03

Suresh Kamrushi · Answer 1 · 2015-11-02T07:17:22.293

0

Might be there is no value in $value. So you can do like this:

public function quote($value = ""){

OR

public function quote($value){
        if(isset($value) && $value != "") {
           $connection = $this -> connect();
           return "'".$connection -> real_escape_string(string ($value))."'";
       } else {
           return false;
       }
    }

edited Nov 02 '15 at 07:17

answered Nov 02 '15 at 07:09

Suresh Kamrushi

15,627
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$value takes input from form. – Shreesh Katyayan Nov 02 '15 at 07:16

score 0 · Answer 2 · answered Nov 02 '15 at 07:10

0

Change string $value to (string) $value

answered Nov 02 '15 at 07:10

tliokos

3,606
3
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28

Tried. Raises notice for conversion from _array_ to _string_. – Shreesh Katyayan Nov 02 '15 at 07:15

score 0 · Answer 3 · answered Nov 02 '15 at 07:11

0

remove string

public function quote($value){
    $connection = $this -> connect();
    return "'".$connection -> real_escape_string($value)."'";
                                                 ^
}

answered Nov 02 '15 at 07:11

Niranjan N Raju

12,047
4
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41

Warning: mysqli::real_escape_string() expects parameter 1 to be string, array given in D:\xampp\htdocs\fizzy\submit.php on line 48 – Shreesh Katyayan Nov 02 '15 at 07:16
`print_r($value);` see what is there inside that. I think according to error, its `array`. – Niranjan N Raju Nov 02 '15 at 07:19
Yeah it gives an array – Shreesh Katyayan Nov 02 '15 at 07:41
@ShreeshKatyayan How can u convert array to string. It will not work. Use `implode()` – Niranjan N Raju Nov 02 '15 at 07:46

score 0 · Answer 4 · answered Nov 02 '15 at 07:17

0

Take a look at last code line

return "'".$connection -> real_escape_string(string $value)."'";

Here parameter passed to function is (string $value).As you are type casting $value,it should be ((string)$value).

answered Nov 02 '15 at 07:17

Dinesh Belkare

639
8
24

Parse error: syntax error T_VARIABLE

4 Answers4