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PHP PDO SQL Server 数据库问题

转载 作者:行者123 更新时间:2023-11-29 13:18:14 25 4
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我使用 SQL Server 作为数据库并使用 PHP PDO 进行连接,创建注册页面时,执行查询时出现此错误

Notice: Array to string conversion in C:\webdev\classes\DB.php on line 36 There was a problem creating an account.PHP Notice: Array to string conversion in C:\webdev\classes\DB.php on line 36

第 36 行 - $this->_query->bindValue($x, $param);

<?php
class DB {
public static $instance = null;

private $_pdo = null,
$_query = null,
$_error = false,
$_results = null,
$_count = 0;

private function __construct() {
try {

$this->_pdo = new PDO('sqlsrv:server=' . Config::get('sqlsrv/servername') . ';database=' . Config::get('sqlsrv/db'), Config::get('sqlsrv/username'), Config::get('sqlsrv/password'));
} catch(PDOExeption $e) {
die($e->getMessage());
}
}

public static function getInstance() {
// Already an instance of this? Return, if not, create.
if(!isset(self::$instance)) {
self::$instance = new DB();
}
return self::$instance;
}

public function query($sql, $params = array()) {

$this->_error = false;

if($this->_query = $this->_pdo->prepare($sql)) {
$x = 1;
if(count($params)) {
foreach($params as $param) {
/* Line 36 */ $this->_query->bindValue($x, $param);
$x++;
}
}

if($this->_query->execute()) {
$this->_results = $this->_query->fetchAll(PDO::FETCH_OBJ);
$this->_count = $this->_query->rowCount();
} else {
$this->_error = true;
}
}

return $this;
}

public function get($table, $where) {
return $this->action('SELECT *', $table, $where);
}

public function delete($table, $where) {
return $this->action('DELETE', $table, $where);
}

public function action($action, $table, $where = array()) {
if(count($where) === 3) {
$operators = array('=', '>', '<', '>=', '<=');

$field = $where[0];
$operator = $where[1];
$value = $where[2];

if(in_array($operator, $operators)) {
$sql = "{$action} FROM {$table} WHERE {$field} {$operator} ?";

if(!$this->query($sql, array($value))->error()) {
return $this;
}

}

return false;
}
}

public function insert($table, $fields = array()) {
$keys = array_keys($fields);
$values = null;
$x = 1;

foreach($fields as $value) {
$values .= "?";
if($x < count($fields)) {
$values .= ', ';
}
$x++;
}

$sql = "INSERT INTO {$table} (`" . implode('`, `', $keys) . "`) VALUES ({$values})";

if(!$this->query($sql, $fields)->error()) {
return true;
}

return false;
}

public function update($table, $id, $fields = array()) {
$set = null;
$x = 1;

foreach($fields as $name => $value) {
$set .= "{$name} = ?";
if($x < count($fields)) {
$set .= ', ';
}
$x++;
}

$sql = "UPDATE users SET {$set} WHERE id = {$id}";

if(!$this->query($sql, $fields)->error()) {
return true;
}

return false;
}

public function results() {
// Return result object
return $this->_results;
}

public function first() {
return $this->_results[0];
}

public function count() {
// Return count
return $this->_count;
}

public function error() {
return $this->_error;
}}

为什么会这样,我使用了相同的代码并且有一个 mysql 数据库,它将数据发送到数据库没有问题,为什么 SQL Server 会出现这种情况?

最佳答案

其中一个 $param 迭代可能以数组形式出现:

if(count($params)) {
foreach($params as $param) {

if(is_array($param) || is_object($param)){ $param=''; }

$this->_query->bindValue($x, $param);
$x++;
}
}

调试建议公共(public)函数insert()

// add a debug parameter
public function insert($table, $fields = array(), $debug = false) {

$return = false;

if(is_array($fields) && count($fields) > 0 && $table != ''){

// build SQL and debug SQL
$sql = "INSERT INTO '$table' (";
$debug_sql = $sql;

// declare variables
$sql_fields = '';
$values = '';
$debug_values = '';

foreach($fields as $k=>$v) {

// encase fields and values in quotes
$sql_fields.= "'$k',";
$values.= "?,";
$debug_values.= "'$v',";
}

// remove trailing commas
$sql_fields = substr($sql_fields, 0, -1);
$values= substr($values, 0, -1);
$debug_values= substr($debug_values, 0, -1);

// finish SQL and debug SQL
$sql.= "$sql_fields) VALUES ($values)";
$debug_sql.= "$sql_fields) VALUES ($debug_values)";

if($debug === true) {
$return = $debug_sql;
}
else {
if(!$this->query($sql, $fields)->error()) {
$return = true;
}
}
}

return $return;
}

// now change the insert call to look like this
die($this->_db->insert('dbo.users', $fields, true)); // <-- notice the true parameter

/**
* Use the output to directly run the SQL from the MSSQL admin console or whatever they call it and it will provide a much more useful error description
*/

关于PHP PDO SQL Server 数据库问题,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/21195483/

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