Question

这是我的PHP文件。我无法从fyp1中选择fyp1.code = $tcode（即我们输入的输入）的截止日期。请帮忙。

<?php
include('inc/db.php');

if (isset($_POST['submit'])) {

$tcode = $_POST['tcode'];
$idno = $_POST['idno'];
$sname = $_POST['sname'];
$datesub = date("Y-m-d");

$sql = "SELECT * FROM fyp1";
$select = mysql_query($sql,"SELECT deadline from fyp1 where fyp1.code = '$tcode'");

$name   = $_FILES['file']['name'];
$tmp_name = $_FILES['file']['tmp_name'];

if ($name && $tcode){
    $Location = "S.File/$name";
    move_uploaded_file($tmp_name, $Location);
    $query = mysql_query("INSERT INTO submission (taskcode,idno,name,file,time,dead) VALUES ('$tcode','$idno','$sname','$name','$datesub','$select')");
    header('Location:DisplayNews.php');
}else
    die("Please select a file");
}

?>

Answer 1

请看一下php devdocs！

mixed mysql_query ( string $query [, resource $link_identifier = NULL ] )

您必须删除以下行...

$sql = "SELECT * FROM fyp1";
$select = mysql_query($sql,"SELECT deadline from fyp1 where fyp1.code = '$tcode'");

并用有效的电话替换它们：

$sql = "SELECT deadline ".
       "FROM fyp1 ".
       "WHERE code = '".mysql_real_escape_string($tcode)."' ");

出于安全原因，请不要直接在SQL语句中插入$ _POST变量！

根据我们的输入从其他表中选择数据

1 个答案: