MySQLiprepared statements 错误报告 [英] MySQLi prepared statements error reporting

问题描述

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' statement captures all errors associated with the complete execution of the query it defines?

Thanks C

解决方案

Each method of mysqli can fail. You should test each return value. If one fails, think about whether it makes sense to continue with an object that is not in the state you expect it to be. (Potentially not in a "safe" state, but I think that's not an issue here.)

Since only the error message for the last operation is stored per connection/statement you might lose information about what caused the error if you continue after something went wrong. You might want to use that information to let the script decide whether to try again (only a temporary issue), change something or to bail out completely (and report a bug). And it makes debugging a lot easier.

$stmt = $mysqli->prepare("INSERT INTO testtable VALUES (?,?,?)");
// prepare() can fail because of syntax errors, missing privileges, ....
if ( false===$stmt ) {
  // and since all the following operations need a valid/ready statement object
  // it doesn't make sense to go on
  // you might want to use a more sophisticated mechanism than die()
  // but's it's only an example
  die('prepare() failed: ' . htmlspecialchars($mysqli->error));
}

$rc = $stmt->bind_param('iii', $x, $y, $z);
// bind_param() can fail because the number of parameter doesn't match the placeholders in the statement
// or there's a type conflict(?), or ....
if ( false===$rc ) {
  // again execute() is useless if you can't bind the parameters. Bail out somehow.
  die('bind_param() failed: ' . htmlspecialchars($stmt->error));
}

$rc = $stmt->execute();
// execute() can fail for various reasons. And may it be as stupid as someone tripping over the network cable
// 2006 "server gone away" is always an option
if ( false===$rc ) {
  die('execute() failed: ' . htmlspecialchars($stmt->error));
}

$stmt->close();

Just a few notes six years later...

The mysqli extension is perfectly capable of reporting operations that result in an (mysqli) error code other than 0 via exceptions, see mysqli_driver::$report_mode.
die() is really, really crude and I wouldn't use it even for examples like this one anymore.
So please, only take away the fact that each and every (mysql) operation can fail for a number of reasons; even if the exact same thing went well a thousand times before....

这篇关于MySQLiprepared statements 错误报告的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！