doufu4333 2018-08-01 01:58
浏览 52

插入当前表并更新到另一个表php

I have two tables named interview and personal_info. I'm having a problem inserting values into my database. I need to insert the values of txtTest, txtIntervieweddBy, txtRemarksDate, txtComment and txtRemarks to interview table. There's also this txtRemarksStatus in the same form but its use is to update a row in a different table named personal_info.

<?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();

?>
  • 写回答

1条回答 默认 最新

  • douxiuyu2028 2018-08-01 07:15
    关注

    Use it Like this:-

     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();
    
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