我开始使用ajaxCRUD,这似乎是一个简单但有效的类。
我有一个这样的表用户:
CREATE TABLE tbl_users (
usr_id INT NOT NULL AUTO_INCREMENT ,
usr_name VARCHAR( 64 ) NOT NULL DEFAULT '' ,
usr_surname VARCHAR( 64 ) NOT NULL DEFAULT '' ,
usr_pwd VARCHAR( 64 ) NOT NULL ,
usr_level INT( 1 ) NOT NULL DEFAULT 0,
PRIMARY KEY ( usr_id )
) ENGINE = InnoDB;
我的 PHP 代码是:
<?php
$tblUsers->omitPrimaryKey();
$tblUsers->displayAs("usr_name", "Nome");
$tblUsers->displayAs("usr_surname", "Cognome");
$tblUsers->displayAs("usr_pwd", "Password");
$tblUsers->displayAs("usr_level", "Livello");
$tblUsers->omitField("usr_pwd");
$tblUsers->modifyFieldWithClass("usr_pwd", "password");
$allowable_vals = array(1 => "USER", 2 => "ADMIN", 3 => "SUPERUSER");
$tblUsers->defineAllowableValues("usr_level", $allowable_vals);
$tblUsers->formatFieldWithFunction('usr_level', 'make_usr_level');
$tblUsers->onAddExecuteCallBackFunction("AddCallBack");
$tblUsers->addButtonToRow("Dettagli", "UserDetails.php");
$tblUsers->addButtonToRow("Collega case", "HousesList.php");
$tblUsers->setLimit(30);
#my self-defined functions used for formatFieldWithFunction
function make_usr_level($val){
return FAUser::LevelToString($val);
}
function AddCallBack($array)
{
//these indexes are the fields of the db
$success = qr("INSERT INTO tbl_users
( usr_id,
usr_name,
usr_surname,
usr_pwd,
usr_level )
VALUES (NULL,
'" . $array[usr_name] . "',
'" . $array[usr_surname] . "',
AES_ENCRYPT('" . $array[usr_pwd] . "',
SHA2('FonteAlma_2013', 512)),
" . $array[usr_level] . ");");
}
#actually show to the table
$tblUsers->showTable();
?>
该表正确显示并显示了已插入的用户,但是当我添加新用户时,我得到:
4 - TheGivenName - TheGivenSurname - [BLOB - 7 B] - 0
所以密码和用户级别都没有设置好。
此外,当我将级别字段编辑到表中时,该表是一个包含三个级别的组合框,无论我选择什么值,我总是得到 0。
我哪里错了?
问候。