Question

我试图在选择查询返回0（未找到原始数据）后插入到多表中选择查询工作和插入查询从未完成时，子网站“displayid”并且没有任何语法错误

代码：

<?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

使用此代码：

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

对更新命令进行类似的更改

Answer 2

你实际上有一个错误

$conn->query($sqlfail)

应该是

$conn->query($sqlfail);

Answer 3

AND stats='$ok'";

我无法看到具有此名称的变量我认为您的意思是AND stats='ok'";

在php中选择查询mysql后运行insert查询

3 个答案: