在PHP中選擇查詢mysql之后運行插入查詢

Question

我嘗試在選擇查詢返回0（未找到原始數據）后選擇表插入多表，但在提交“ displayid”且沒有任何語法錯誤時插入查詢從未完成

碼：

<?php
    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)
        }
    }
?>

Answer 1

使用此代碼：

  $sqlfail = "INSERT INTO  fail(fname,lname,tname,funame,idnum,m_phone,reg_date)
VALUES ('".$pfname."','".$plname."','".$ptname."','".$pfuname."','".$pidnum."','".$pm_phone."','".$todaydate."')";

對update命令也進行類似的更改

Answer 2

你實際上有一個錯誤

$conn->query($sqlfail) 

應該

$conn->query($sqlfail);  

Answer 3

AND stats='$ok'";

我看不到具有此名稱的變量，我認為您的意思是AND stats='ok'";