我的 PHP 脚本出现问题,从 MySQL 调用的值作为字符串返回,尽管在数据库中被标记为 int
和 tinyint
.
这是一个问题,因为在将基于 MySQL 日期的数组转换为 JSON 数据时,本应为整数的值会被放在双引号中,这会在使用该 JSON 数据的 Javascript 和 iPhone 应用程序中造成问题。我得到的 JSON 值看起来像 "key" : "1"
,当我想要的是"key" : 1
.
经过一些研究,似乎it should be possible to get the values as their native type只要有 PHP 5.3 和 mysqlnd
安装的模块。我有 5.3.3 和 phpinfo()
似乎表明我有 mysqlnd
模块安装并运行:
mysqlnd enabled
Version mysqlnd 5.0.10 - 20111026
但是,我的值仍以字符串形式返回。
我看过 PHP manual entry for mysqlnd ,而且我总是有可能遗漏明显的东西,但我没有看到任何表明我需要在代码中执行任何特定操作才能获取 native 值的信息。
我究竟该怎么做才能让 PHP 中的 MySQL 函数以其 native 类型提供 MySQL 结果?
为了使下面的答案更加有趣,这是我用来连接到数据库的命令:
private function databaseConnect()
{
$this->mysqli = new mysqli(Database::$DB_SERVER, Database::$DB_USERNAME, Database::$DB_PASSWORD);
$this->mysqli->set_charset("utf8");
return true;
}
private function dbConnect()
{
Database::$USE_MYSQLI = extension_loaded('mysqli');
if (!$this->databaseConnect())
{
echo "Cannot Connect To The Database Server";
throw new Exception();
}
if (!$this->databaseSelectDB())
{
echo "The database server connected, but the system could not find the right database";
throw new Exception();
}
}
private function databaseQuery($query)
{
return $this->mysqli->query($query);
}
public function doQuery($query)
{
$result = $this->databaseQuery($query);
if ($result == FALSE)
{
//ErrorHandler::backtrace();
die("This query did not work: $query");
}
return $result;
}
private function getRows($table, $matches, $orderBy = array(), $limit = array())
{
$calcFoundRows = '';
if (count($limit) > 0)
{
$calcFoundRows = ' SQL_CALC_FOUND_ROWS';
}
$query = 'SELECT ' . $calcFoundRows . ' * FROM ' . $table;
if (count($matches) > 0)
{
$query .= ' WHERE ';
$keys = array_keys($matches);
$first = true;
foreach ($keys as $key)
{
if (!$first)
{
$query .= ' AND ';
}
$first = false;
// now he is safe to add to the query
// the only time this is an array is when this is called by getSelectedUsers or getSelectedArticles
// in that case it is an array of array's as the key (which is the column name) may have more than
// one condition
if (is_array($matches[$key]))
{
$firstForColumn = true;
foreach ($matches[$key] as $conditions)
{
if (!$firstForColumn)
{
$query .= ' AND ';
}
$firstForColumn = false;
// if the value is an array we generate an OR selection
if (is_array($conditions[1]))
{
$firstOr = true;
$query .= '(';
foreach ($conditions[1] as $value)
{
if (!$firstOr)
{
$query .= ' OR ';
}
$firstOr = false;
// clean this guy before putting him into the query
$this->cleanMySQLData($value);
if ($conditions[0] == Selection::$CONTAINS)
{
//$query .= 'MATCH (' . $key . ') AGAINST (' . $value . ') ';
$value = trim($value, "'");
$value = "'%" . $value . "%'";
$query .= $key . ' LIKE ' . $value;
}
else
{
$query .= $key . ' ' . $conditions[0] . ' ' . $value;
}
}
$query .= ')';
}
else
{
// clean this guy before putting him into the query
$var = $conditions[1];
$this->cleanMySQLData($var);
if ($conditions[0] == Selection::$CONTAINS)
{
//$query .= 'MATCH (' . $key . ') AGAINST (' . $var . ') ';
$var = trim($var, "'");
$var = "'%" . $var . "%'";
$query .= $key . ' LIKE ' . $var;
}
else
{
$query .= $key . ' ' . $conditions[0] . ' ' . $var;
}
}
}
}
else
{
// clean this guy before putting him into the query
$this->cleanMySQLData($matches[$key]);
$query .= $key . " = " . $matches[$key];
}
}
}
if (count($orderBy) > 0)
{
$query .= " ORDER BY ";
$first = true;
foreach ($orderBy as $orderCol)
{
if (!$first)
{
$query .= ',';
}
$query .= $orderCol;
$first = false;
}
}
if (count($limit) > 0)
{
$query .= ' LIMIT ' . $limit[0];
if (count($limit) > 1)
{
$query .= ',' . $limit[1];
}
}
$result = $this->doQuery($query);
$data = array();
while ($row = $this->databaseFetchAssoc($result))
{
$data[] = $row;
}
if (strlen($calcFoundRows) > 0)
{
$numRows = $this->databaseCountFoundRows();
$key = '^^' . $table . '_selectionCount';
Session::getSession()->putUserSubstitution($key, $numRows);
}
return $data;
}
最佳答案
What exactly do I do to get my MySQL functions in PHP to give me the MySQL results in their native type?
连接到数据库,然后准备查询、执行查询、绑定(bind)结果,然后获取结果。
让我们逐行执行这些步骤:
$conn = new Mysqli('localhost', 'testuser', 'test', 'test');
$stmt = $conn->prepare("SELECT id FROM config LIMIT 1");
$stmt->execute();
$stmt->bind_result($id);
$stmt->fetch();
var_dump($id); # it's an int!
这对我有用。随着您编写的代码越来越复杂,您将需要定位查询数据库的位置。检查您是否正在使用 Mysqli::prepare()
如果没有,请介绍。
您还需要使用 Mysqli_Stmt::execute()
然后 Mysqli_Stmt::bind_result()
否则,不会为该结果列保留(此处为整数)类型。
关于php - 我如何确保来自 MySQL 的值在 PHP 中保持其类型?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/14180981/