如何确保MySQL中的值保持其在PHP中的类型？
作者：互联网
我在我的PHP脚本中遇到一个问题,即MySQL中调用的值作为字符串返回,尽管在数据库中标记为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,应该是整数的值放在双引号中,这会导致使用该JSON数据的Javascript和iPhone应用程序出现问题.我得到的JSON值看起来像“key”：“1”,当我想要的是“key”时：1.
经过一些研究,似乎只要一个人拥有PHP 5.3,就安装了mysqlnd模块.我有5.3.3和phpinfo()似乎表明我已经安装并运行了mysqlnd模块：
mysqlnd enabled
Version mysqlnd 5.0.10 - 20111026
但是,我的值仍然以字符串形式返回.
我已经看过了PHP manual entry for mysqlnd,而且总是有可能我错过了显而易见的东西,但我没有看到任何迹象表明我需要在我的代码中做任何特定的事情才能获得原生值.
我究竟该怎么做才能在PHP中获取我的MySQL函数,以便为我的原生类型提供MySQL结果？
为了说明下面的答案,这是我用来连接数据库的命令：
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;
}
解决方法:
What exactly do I do to get my MySQL functions in PHP to give me the MySQL results in their native type?
您连接到数据库,然后准备您的查询,执行它,绑定结果,然后您获取它.
让我们逐行完成这些步骤：
$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!
这适合我.在编写代码时更复杂,您需要找到查询数据库的位置.检查您是否正在使用Mysqli::prepare(),如果没有,请介绍它.
您还需要使用Mysqli_Stmt::execute()然后使用Mysqli_Stmt::bind_result(),否则不会为该结果列保留(此处为整数)类型.
标签：php,mysql,types,casting,mysqlnd 来源： https://codeday.me/bug/20190625/1286573.html