Question

这是从html页面中的表单触发的脚本。

<?php
// title , notice , to , from , subject . time
 // var: flag1, flag2 ;
require('check_session.php');
require('db_init_.php');

$emailid = $_SESSION["emailid"];
$title = $_POST["title"];
$notice = $_POST["notice"];
if(empty($_POST["firstyear"])) {$firstyear="";} 
else 
 {  $firstyear = "";
  foreach ($_POST['firstyear'] as $s)
 {
     $firstyear = $firstyear." , " .$s; // setting Selected Value
 }
 }
 if(empty($_POST["secondyear"])) {$secondyear="";} 
    else 
   {  $secondyear = "";
   foreach ($_POST['secondyear'] as $s)
   {
     $secondyear = $secondyear ." , ".$s; // setting Selected Value
   }
    }
   if(empty($_POST["thirdyear"])) {$thirdyear="";} 
   else 
     {  $thirdyear = "";
  foreach ($_POST['thirdyear'] as $s)
    {
     $thirdyear = $thirdyear ." , ".$s; // setting Selected Value
  }
   }
   if(empty($_POST["fourthyear"])) {$fourthyear="";} 
   else 
   {  $fourthyear = "";
  foreach ($_POST['fourthyear'] as $s)
    {
     $fourthyear = $fourthyear ." , ".$s; // setting Selected Value
    }
  }


    $to = $firstyear.$secondyear.$thirdyear.$fourthyear;
     $subject = $_POST["subject"];


     echo $emailid . $title .$notice . $to. $subject ;

     $table="accounts_faculty";
     $sql_4="SELECT * FROM $table WHERE emailid='$emailid' ";
      $result4=mysqli_query($conn,$sql_4);

     $row4 = mysqli_fetch_row($result4);
     $first = $row4[1];
     $last = $row4[2];
     $name = $first." ".$last;

     echo $name ;
     $count4=mysqli_num_rows($result4);//faculty

     if($count4 == 1) {$flag1 = 1;}
    else $flag1 = 0;
     if($flag1 == 1)
    {       echo " in flg 1 ";



    $table = "notices";
    $sql_5= "INSERT INTO `minor_naps`.'$table' (`title`, `notice`,                        `to`, `from`, `subject`, `time`)
             VALUES ('$title', '$notice', '$to', '$name', '$subject',           CURRENT_TIMESTAMP);";

     $result5=mysqli_query($conn, $sql_5);   // to be continued
    //var_dump($result5);
    if($result5==1)
    {$flag2= 1;
   // header("location:main-page.php");
    echo 1;
    }

    else {
        $flag2 = 0;
       // header("location:create_notice.html");

    }
    }
      echo $flag1.",".$flag2;

      ?>

check_session.php检查用户会话是否已设置。和db_init_.php是

<?php 
  //database details
   $host = "localhost";
     $username = "root";
   $password = "";
   $db="minor_naps";
   //$table="accounts";


   // Create connection
    $conn = new mysqli($host, $username, $password,$db);

    // Check connection
   if ($conn->connect_error) {
   die("Connection failed: " . $conn->connect_error);
      }
    else {
    # code...

      //echo "<h1>db initialised and Connected successfully</h1>";
       }

  ?>

从html收到数据时，它会从accounts_faculty表中提取数据以获取其名称。然后将其插入表格通知中。

我在此代码中有标记。运行后，没有语法错误，标志1为1但标志2仍为0。那就是数据没有插入表格通知中。

我错过了什么吗？这里有什么错误？以及改进此代码的任何建议吗？

值没有从php脚本插入到MySQL表中

0 个答案: