MySQLi prepared statements error reporting [duplicate]

This question already has an answer here: What to do with mysqli problems? Errors like mysqli_fetch_array(): Argument #1 must be of type mysqli_result and such (1 answer) Closed 5 years ago.

I'm trying to get my head around MySQli and I'm confused by the error reporting. I am using the return value of the MySQLi 'prepare' statement to detect errors when executing SQL, like this:

$stmt_test =  $mysqliDatabaseConnection->stmt_init();
if($stmt_test->prepare("INSERT INTO testtable VALUES (23,44,56)"))
{
 $stmt_test->execute();
 $stmt_test->close();
}
else echo("Statement failed: ". $stmt_test->error . "<br>");

But, is the return value of the prepare statement only detecting if there is an error in the preperation of the SQL statement and not detecting execution errors? If so should I therefore change my execute line to flag errors as well like this:

if($stmt_test->execute()) $errorflag=true;

And then just to be safe should I also do the following after the statement has executed:

if($stmt_test->errno) {$errorflag=true;}

...Or was I OK to start with and the return value on the MySQLi prepare' statem开发者_Go百科ent captures all errors associated with the complete execution of the query it defines?

Thanks C

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Each method of mysqli can fail. Luckily, nowadays mysqli can report every problem to you, all you need is ask. Simply add this single line to the connection code,

mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);

And after that every error will reveal itself. No need to test any return values ever, just write your statements right away:

$stmt = $mysqli->prepare("INSERT INTO testtable VALUES (?,?,?)");
$stmt->bind_param('iii', $x, $y, $z);
$stmt->execute();

When the error occurs at any step, it will throw a usual PHP Exception that can be handled or just reported the same way as any other PHP error. Just make sure you configured PHP error reporting properly, i.e. on the dev server errors are displayed on-screen and on the production server errors are never displayed but logged instead.

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Completeness

You need to check both $mysqli and $statement. If they are false, you need to output $mysqli->error or $statement->error respectively.

Efficiency

For simple scripts that may terminate, I use simple one-liners that trigger a PHP error with the message. For a more complex application, an error warning system should be activated instead, for example by throwing an exception.

Usage example 1: Simple script

# This is in a simple command line script
$mysqli = new mysqli('localhost', 'buzUser', 'buzPassword');
$q = "UPDATE foo SET bar=1";
($statement = $mysqli->prepare($q)) or trigger_error($mysqli->error, E_USER_ERROR);
$statement->execute() or trigger_error($statement->error, E_USER_ERROR);

Usage example 2: Application

# This is part of an application
class FuzDatabaseException extends Exception {
}

class Foo {
  public $mysqli;
  public function __construct(mysqli $mysqli) {
    $this->mysqli = $mysqli;
  }
  public function updateBar() {
    $q = "UPDATE foo SET bar=1";
    $statement = $this->mysqli->prepare($q);
    if (!$statement) {
      throw new FuzDatabaseException($mysqli->error);
    }

    if (!$statement->execute()) {
      throw new FuzDatabaseException($statement->error);
    }
  }
}

$foo = new Foo(new mysqli('localhost','buzUser','buzPassword'));
try {
  $foo->updateBar();
} catch (FuzDatabaseException $e)
  $msg = $e->getMessage();
  // Now send warning emails, write log
}

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Not sure if this answers your question or not. Sorry if not

To get the error reported from the mysql database about your query you need to use your connection object as the focus.

so:

echo $mysqliDatabaseConnection->error

would echo the error being sent from mysql about your query.

Hope that helps