这是我进行更新查询的方式,查询正在运行,但没有成功,也没有错误结果。虽然我已经使用参数手动测试了 MySql 工作台中的查询,但一切正常。我缺少什么。我也不知道为什么如果它不成功,它不会抛出任何异常或错误。
function UpdateLeaveDetailRequestStatus(){
global $connPDO;
date_default_timezone_set("Asia/Karachi");
//$connPDO->beginTransaction();
$currentDate = date('Y-m-d H:i:s');
$binds = array(
":leave_start_date" => $_POST["leaveStartDate"],
":leave_end_date" => $_POST["leaveEndDate"],
":leave_total_days" => $_POST["leaveTotalDays"],
":leave_type" => $_POST["leaveType"],
":leave_in_out_time" => $_POST["leaveInOutTime"],
":leave_duration" => $_POST["leaveDuration"],
":leave_reason" => $_POST["leaveReason"],
":leave_current_status_id" => $_POST["leaveCurrentStatusId"],
":leave_current_status_set_by" => $_COOKIE["userID"],
":leave_current_status_set_dateTime" => $currentDate,
":leave_detail_id" => $_POST["leaveDetailId"]
);
echo $_POST["empId"] ."\n";
echo $_POST["leaveStartDate"]. "\n";
echo $_POST["leaveEndDate"]. "\n";
echo $_POST["leaveTotalDays"]. "\n";
echo $_POST["leaveType"]. "\n";
echo $_POST["leaveInOutTime"]. "\n";
echo $_POST["leaveDuration"]. "\n";
echo $_POST["leaveReason"]. "\n";
echo $_POST["leaveCurrentStatusId"]. "\n";
echo $_COOKIE["userID"]. "\n";
echo $currentDate. "\n";
echo $_POST["leaveDetailId"]. "\n";
}
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