我试图更新i表,但是当我运行PHP脚本时,我没有收到任何错误,也没有任何更新
$link = mysqli_connect("XXX.X.X.X", $username, $password, $database);
if(isset($_REQUEST["UserID"]))
{
$userID = $_REQUEST["UserID"];
}
if(isset($_REQUEST["ClientID"]))
{
$clientID = $_REQUEST["ClientID"];
}
if(isset($_REQUEST["type"]))
{
$type = $_REQUEST["type"];
}
if(isset($_REQUEST["sent"]))
{
$text = isset($_REQUEST["sent"]);
}
$volunadd = "UPDATE CHAT SET VOLUNTEER_TEXT = CONCAT(?,VOLUNTEER_TEXT) WHERE PATIENT_ID = ? AND VOLUNTEER_ID = ?";
$patadd = "UPDATE CHAT SET PATIENT_TEXT = CONCAT(?,PATIENT_TEXT) WHERE PATIENT_ID = ? AND VOLUNTEER_ID = ?";
if($type == "V")
{
$stmt = mysqli_stmt_init($link);
if(mysqli_stmt_prepare($stmt,$volunadd))
{
mysqli_stmt_bind_param($stmt,"sii",$text,$clientID,$userID);
mysqli_stmt_execute($stmt);
}
}
else if($type == "P")
{
$stmt = mysqli_stmt_init($link);
if(mysqli_stmt_prepare($stmt,$patadd))
{
mysqli_stmt_bind_param($stmt,"sii",$text,$userID,$clientID);
mysqli_stmt_execute($stmt);
}
}