Why I am getting Cannot pass parameter 2 by reference error when I am using bindParam with a constant value?

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I'm using this code and I'm beyond frustration:

try {
    $dbh = new PDO('mysql:dbname=' . DB . ';host=' . HOST, USER, PASS);
    $dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
    $dbh->setAttribute(PDO::MYSQL_ATTR_INIT_COMMAND, "SET NAMES 'utf8'");
}
catch(PDOException $e)
{
    ...
}
$stmt = $dbh->prepare('INSERT INTO table(v1, v2, ...) VALUES(:v1, :v2, ...)');
$stmt->bindParam(':v1', PDO::PARAM_NULL); // --> Here's the problem

PDO::PARAM_NULL, null, '', all of them fail and throw this error:

Fatal error: Cannot pass parameter 2 by reference in /opt/...

Answers

You need to use bindValue, not bindParam

bindParam takes a variable by reference, and doesn't pull in a value at the time of calling bindParam. I found this in a comment on the PHP docs:

bindValue(':param', null, PDO::PARAM_INT);

P.S. You may be tempted to do this bindValue(':param', null, PDO::PARAM_NULL); but it did not work for everybody (thank you Will Shaver for reporting.)

Saturday, November 5, 2022

jamey_mcelveen-850ada8b7028

The following works for me:

<?php

$pdo = new PDO("mysql:host=localhost;dbname=test", "root", "pass");
$pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$pdo->setAttribute(PDO::ATTR_EMULATE_PREPARES, false);

$stmt = $pdo->prepare("INSERT INTO `null_test` (`can_be_null`) VALUES (:null)");
$stmt->bindValue(":null", null, PDO::PARAM_NULL);

$stmt->execute();

Pass in PHP's null, with type of PDO::PARAM_NULL. Also, make sure your prepare emulation is set to false. That might help.

Sunday, August 7, 2022

mohammad_reza_khahani

See my notes inline with the code:

    try{  
         $req = $prepared_insertQry_toUniqueRefTable -> execute(array(something));

         // this never executes because an Exception halts it here
         /*if($req == 0){
               $prepared_insertQry_toDuplicateRefTable->execute(array(something));
         }*/
     }
    catch(PDOException $e){
       // this catch grabs the exception and executes the code within instead
       // that code might log an error, echo the error message, or perform
       // alternative logic. In your case you want to execute alterntative logic
       // ie. your query
       $prepared_insertQry_toDuplicateRefTable->execute(array(something));

   }

Thursday, December 22, 2022

turbomanolo

The error means that the 2nd argument is expected to be a reference to a variable.

Since you are not handing a variable but an integer of value 0, it generates said error.

To circumvent this do:

$a = 0;
$update->bind_param("is", $a, $selectedDate);  //LINE 13

In case you want to understand what is happening, as opposed to just fixing your Fatal error, read this: http://php.net/manual/en/language.references.pass.php

Tuesday, November 1, 2022

thakur

It expects the second paramter to be a variable which can be passed by reference. Assuming $stmt1is a PDO statement then, as the docs for bindparam say

Unlike PDOStatement::bindValue(), the variable is bound as a reference and will only be evaluated at the time that PDOStatement::execute() is called.

Your second param is an expression ($_SESSION['quantity'.$i] * $_SESSION['price'.$i]) not a variable. Since you appear to want to evaluate the exptression now, I guess you should used bindValue() instead.

Tuesday, December 6, 2022

brobin

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Why I am getting Cannot pass parameter 2 by reference error when I am using bindParam with a constant value?

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php,mysql,pdo,null,sql-insert