表中的记录未更新

时间:2020-07-01 12:12:17

标签: php mysqli

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);
    }
}

当我在终端中运行查询时,它起作用了Here is the layout of the table 代码中是否有错误?还是应该以其他方式做

0 个答案:

没有答案
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