Parse error: syntax error T_VARIABLE [duplicate]

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
– Sean, Commented Nov 2, 2015 at 7:19
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
– Shreesh Katyayan, Commented Nov 2, 2015 at 7:38
Sorry, error was in function calls and not the function definition. Mark closed :) Thanks for help anyways — Shreesh Katyayan
– Shreesh Katyayan, Commented Nov 2, 2015 at 8:03

tliokos · Accepted Answer · 2015-11-02 07:10:07Z

0

Change string $value to (string) $value

answered Nov 2, 2015 at 7:10

tliokos

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1 Comment

Shreesh Katyayan Over a year ago

Tried. Raises notice for conversion from array to string.

Niranjan N Raju · Accepted Answer · 2015-11-02 07:11:53Z

0

remove string

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

answered Nov 2, 2015 at 7:11

Niranjan N Raju

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4 Comments

Shreesh Katyayan Over a year ago

Warning: mysqli::real_escape_string() expects parameter 1 to be string, array given in D:\xampp\htdocs\fizzy\submit.php on line 48

Niranjan N Raju Over a year ago

print_r($value); see what is there inside that. I think according to error, its array.

Shreesh Katyayan Over a year ago

Yeah it gives an array

Niranjan N Raju Over a year ago

@ShreeshKatyayan How can u convert array to string. It will not work. Use implode()

Dinesh Belkare · Accepted Answer · 2015-11-02 07:17:07Z

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 2, 2015 at 7:17

Dinesh Belkare

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Comments

Suresh Kamrushi · Accepted Answer · 2015-11-02 07:17:22Z

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 2, 2015 at 7:17

answered Nov 2, 2015 at 7:09

Suresh Kamrushi

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1 Comment

Shreesh Katyayan Over a year ago

$value takes input from form.

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