php - 在php中选择查询mysql之后运行更新或插入查询

我已经花了超过24小时试图在选择查询后运行更新或插入查询，但是在提交“ displayid”时选择了已完成的查询并从未完成更新或插入查询

码＃＃

if($_POST["displayid"]==TRUE) {

    $sqlid = "SELECT * FROM doc1 WHERE idnum ='$pidnum' AND stats='$ok'";
    $result = mysqli_query($conn, $sqlid);
    if (mysqli_num_rows($result) > 0) {
         $sqlup = "UPDATE doc1 SET  m_phone='$pmphone', seen='$dataseen' WHERE  idnum ='$pidnum'";
        mysqli_query($conn, $sqlup);
        $found=1;
    }
    else {
        $found=0;
        $sqlfail="INSERT INTO fail(fname,lname,tname,funame,idnum,m_phone,reg_date) VALUES('$pfname','$plname','$ptname','$pfuname','$pidnum','$pm_phone','$todaydate')";
        $conn->query($sqlfail);
    }

}

最佳答案

首先您更新查询是错误的。
为了检查错误，请添加

error_reporting(E_ALL);
ini_set('display_errors', 1);

更新的代码

if ($_POST["displayid"] == TRUE) {

    $sqlid = "SELECT * FROM doc1 WHERE idnum ='$pidnum' AND stats='$ok'";
    $result = mysqli_query($conn, $sqlid);
    if (mysqli_num_rows($result) > 0) {
        $sqlup = "UPDATE doc1 SET  m_phone='$pm_phone', seen='$dataseen' WHERE  idnum ='$pidnum'";
        mysqli_query($conn, $sqlup);
        $found = 1;
    } else {
        $found = 0;
        $sqlfail = "INSERT INTO fail(fname,lname,tname,funame,idnum,m_phone,reg_date)
VALUES ('$pfname','$plname','$ptname','$pfuname','$pidnum','$pm_phone','$todaydate')";
        $conn->query($sqlfail);
    }
}