在 PHP 中选择查询 MySQL 后运行更新或插入查询


run update or insert query after select query mysql in php

>我花了超过 24 小时尝试在选择查询后运行更新或插入查询,但选择查询完成并在提交"displayID"时更新或插入从未完成的查询

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) {
         $sqlup = "UPDATE doc1 SET  m_phone='$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);
    }
}

首先,您更新查询是错误的。要检查错误,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);
    }
}

$conn对象来自哪里? 试试这个。.

<?php
if($_POST["displayid"])
{
$sqlid = "SELECT * FROM doc1 WHERE idnum ='$pidnum' AND stats='$ok'";
$result = mysqli_query($conn, $sqlid);
if (mysqli_num_rows($result) > 0)
{
   $sqlup= "UPDATE doc1 SET  m_phone='$pm_phone' AND 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')";
mysqli_query($conn, $sqlfail);

}
}