我有以下行来设置我的sql cmd。
$mySqlCmd = "sqlcmd -S $server -U $username -P $pwd -d $dbname -o $lis -i $sqlScript -v fileDate = $fileDate, filePath = $filePath, sqlLoadErrors = $sqlLoadErrorPath"
Invoke-Expression $mySqlCmd
这似乎不适用于多个变量。如果删除我尝试传递给SQL脚本的最后两个变量,SQL脚本将运行并运行。
在我的SQL脚本上,我有以下内容:
DECLARE
@currentDate NVARCHAR(25),
@filePath NVARCHAR(25),
@sqlLoadErrors NVARCHAR(25)
SET @currentDate = $(fileDate)
SET @filePath = $(filePath)
SET @sqlLoadErrors = $(sqlLoadErrors)
在PowerShell脚本中是否存在将变量传递给SQL脚本的错误?
答案 0 :(得分:0)
试试这个:
public function __construct() {
// Assign the CodeIgniter super-object
$this->CI = & get_instance();
}
//Now for loading a model
public function addAudit($id_user,$description){
$this->CI->load->model('your_model_name');
$response_audit = $this->CI->your_model_name->addAudit($id_user,$description);
if($response_audit){
log_message('debug',' Product --addAudit :: Response received from model');
}else{
log_message('debug',' Product --addAudit :: Response didnot received from model');
}
}