Question

我有两个表，分别是“采访”和“ personal_info”。我在将值插入数据库时遇到问题。我需要将txtTest，txtIntervieweddBy，txtRemarksDate，txtComment和txtRemarks的值插入到采访表中。 txtRemarksStatus的格式也相同，但是它的用途是更新另一个名为personal_info的表中的行。

<?php
    include_once 'php/connection.php';
    $connect = mysqli_connect("localhost", "root", "", "trackingsystemdb");
    $number = count($_POST["txtIntervieweddBy"]);

    if($number > 0)
    {
        for($i=0; $i<$number; $i++)
        {
            if(trim($_POST["txtIntervieweddBy"][$i] != ''))
            {
                $sql = "INSERT INTO interview (applicant_code, interviewed, dateInterviewed, comment, remarks) VALUES('".mysqli_real_escape_string($connect, $_POST["txtTest"][$i])."','".mysqli_real_escape_string($connect, $_POST["txtIntervieweddBy"][$i])."','".mysqli_real_escape_string($connect, $_POST["txtRemarksDate"][$i])."','".mysqli_real_escape_string($connect, $_POST["txtComment"][$i])."','".mysqli_real_escape_string($connect, $_POST["txtRemarks"][$i])."')";

            }
        }
    }

    $id = $_GET['id'];
    $txtRemarksStatus=$_GET['txtRemarksStatus'];
    $sql = "UPDATE personal_info SET status =  '$txtRemarksStatus'          
            WHERE applicant_code = '$id'";

    if (mysqli_multi_query($connect, $sql)) {
    } else {
        echo "Error ".mysqli_error($connect);
    }

    header('Location: addinterview.php?insert=sucess');
    exit();

?>

Answer 1

像这样使用它：-

 include_once 'php/connection.php';
    $connect = mysqli_connect("localhost", "root", "", "trackingsystemdb");
    $number = count($_POST["txtIntervieweddBy"]);

    if($number > 0)
    {
        for($i=0; $i<$number; $i++)
        {
            if(trim($_POST["txtIntervieweddBy"][$i] != ''))
            {
                $sql = "INSERT INTO interview (applicant_code, interviewed, dateInterviewed, comment, remarks) VALUES('".mysqli_real_escape_string($connect, $_POST["txtTest"][$i])."','".mysqli_real_escape_string($connect, $_POST["txtIntervieweddBy"][$i])."','".mysqli_real_escape_string($connect, $_POST["txtRemarksDate"][$i])."','".mysqli_real_escape_string($connect, $_POST["txtComment"][$i])."','".mysqli_real_escape_string($connect, $_POST["txtRemarks"][$i])."')";
mysqli_query($connect, $sql);
            }
        }
    }

    $id = $_GET['id'];
    $txtRemarksStatus=$_GET['txtRemarksStatus'];
    $sql = "UPDATE personal_info SET status =  '$txtRemarksStatus'          
            WHERE applicant_code = '$id'";

    if (mysqli_query($connect, $sql)) {
    } else {
        echo "Error ".mysqli_error($connect);
    }

    header('Location: addinterview.php?insert=sucess');
    exit();

插入到当前表并更新到另一个表php

1 个答案: