Worker<MyDTO> q = new Worker<MyDTO>();
我所有的条件都得到满足。我什至在回显中获取了所有正确的值，但只有第一个插入查询（即if (isset($_POST["AddErrorCode"]))
{
$AddErrorCodeDB = $_POST["AddErrorCode"];
$AddErrorDescriptionDB = $_POST["AddErrorDescription"];
$AddQuantityDB = $_POST["AddQuantity"];
$AddStartDateDB = $_POST["AddStartDate"];
$AddCompletionDateDB = $_POST["AddCompletionDate"];
$AddReviewTypeDB = $_POST["AddReviewType"];
session_start();
$WO_ID = $_SESSION['SELECTED_WO_ID'];
if ($AddReviewTypeDB === 'PR')
{
$AddReviewerType = 'Peer Review';
$insert = "INSERT INTO `wo_errorinfo` (
`Error_Code` ,
`Error_Description` ,
`Error_Quantity` ,
`Review_Type` ,
`WO_NO`) VALUES (
'$AddErrorCodeDB' ,
'$AddErrorDescriptionDB' ,
'$AddQuantityDB' ,
'$AddReviewerType' ,
'$WO_ID')";
if ($AddCompletionDateDB === '')
{
//echo 'ritwik';
$status = 'Peer RWK';
$update = "UPDATE `associated_wos` SET `WO Status` = '$status' WHERE `ID` = '$WO_ID'";
}
else
{
//echo 'ritwik1';
$status = 'Peer Review Complete';
$update = "UPDATE `associated_wos` SET `WO Status` = '$status' WHERE `ID` = '$WO_ID'";
}
$sql = "SELECT * FROM `wo_reviewerqa` WHERE `WO_ID` = '$WO_ID' AND `reviewType` = '$AddReviewerType'";
$result = mysqli_query($conn, $sql);
$num_rows = mysqli_num_rows($result);
//echo $num_rows;
if ($num_rows === 0)
{
//echo 'ritwik';
$insertreview = "INSERT INTO `wo_reviewerqa` (
`reviewType`,
`reviewStartDate`,
`reviewCompleteDate`,
`WO_ID`) VALUES (
'$AddReviewerType',
'$AddStartDateDB',
'$AddCompletionDateDB' ,
'$WO_ID')";
//echo $insertreview;
}
else
{
if ($AddStartDateDB !== '')
{
echo "<script type='text/javascript'>alert('Review Already Started, Start Date cant be changed');</script>";
}
}
if($conn->query($insertreview) === True)
{
echo "<script type='text/javascript'>alert('Start date updated successfully');</script>";
}
if ($conn->query($insert) === True)
{
echo "<script type='text/javascript'>alert('Error Code Submitted successfully');</script>";
}
}
在工作），而所有其他查询都对表没有影响。我们不能在一个会话期间插入多个表中。是由于insert into 'wo_errorinfo'吗？我现在已经尝试解决这一问题超过1天，但无法解决。
答案 0 :(得分：2)
您需要执行语句，当前仅执行$sql。
您还应该避免通过连接字符串来构建查询，因为这会使您容易受到SQL注入攻击的影响，在这种情况下，用户可以通过在输入中传递特殊字符来修改查询。您应该使用mysqli::prepare，例如：
if ($stmt = $mysqli->prepare("SELECT District FROM City WHERE Name=?")) {
/* bind parameters for markers */
$stmt->bind_param("s", $city);
/* execute query */
$stmt->execute();
}