PHP:如何在mysqli提交期间检查错误？

Question

我使用mysqli commit语句插入了大量记录,使用来自http://www.php.net/manual/en/mysqli.commit.php#88857的代码.然后我使用:$ mysqli-> affected_rows检查受影响的行,但即使已插入所有记录,我也会得到零影响的行.

如何检查提交失败并检索错误？谢谢

Answer 1

你可以这样做:

mysqli_autocommit($dbconn, FALSE);

$errors = array();

if (!$mysqli->query(/* some SQL query */)) {
    $errors[] = $mysqli->error;
}
// ... more queries like the above
if(count($errors) === 0) {
    $mysqli->commit()
} else {
    $mysqli->rollback();
    print_r($errors);
}

当查询出错时,它会将错误添加到$ errors数组中,这样您就会知道出了什么问题.您还可以添加带有查询标识符的键,以便了解哪个查询出错.

为了更好地处理,您可以为此编写一个UnitOfWork类:

class UnitOfWork
{
    protected $_db;
    protected $_errors;
    protected $_queries;
    public function __construct($db) {
        $this->_db = $db;
        $this->_errors = array();
        $this->_queries = array();
    }
    public function addQuery($id, $sql) {
        $this->_queries[$id] = $sql;
        return $this;
    }
    public function getErrors() {
        return $this->_errors;
    }    
    public function try() {
        $this->_db->autocommit($this->_db, FALSE);
        foreach($this->_queries as $id => $query) {
            if ($this->_db->query($query) === FALSE) {
                $this->_errors[$id] = $this->_db->error;
            }
        }
        $hasErrors = count($this->_errors);
        ($hasErrors) ? $this->_db->rollback() : $this->_db->commit();
        $this->_db->autocommit($this->_db, TRUE);
        return !$hasErrors; // return true on success
    }
}

你可以像使用它一样

$unit = new UnitOfWork($mysqli);
$unit->addQuery('foo', 'SELECT foo FROM somewhere')
     ->addQuery('bar', 'SELECT bar FROM somewhereElse')
     ->addQuery('baz', 'SELECT baz WITH brokenQuery');

if($unit->try() === FALSE) {
    print_r($unit->getErrors());
}