将数据插入SQL表时出错

-1

我在尝试从html表单插入数据时收到此错误。将数据插入SQL表时出错

Error: INSERT INTO uren (aantaluren, projectname, datum) VALUES ('6','dropdown','2017-02-10' 
You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near '' at line 2

这是我的，我想在添加数据的页面代码：

<?php 
session_start(); 
require_once(dirname(__FILE__)."/simpleusers/su.inc.php"); 

$mysqli = mysqli_connect("localhost","root","","urenregistratie"); 
$sqlSelect="SELECT name, projectId FROM projecten"; 
$result = $mysqli -> query ($sqlSelect); 

while ($row = mysqli_fetch_array($result)) { 
echo "option value='" . $row['name'] . "'>" . $row['projectId'] . "</option>"; 
} 

$SimpleUsers = new SimpleUsers(); 

// This is a simple way of validating if a user is logged in or not. 
// If the user is logged in, the value is (bool)true - otherwise (bool)false. 
if(!$SimpleUsers->logged_in) 
{ 
    header("Location: login.php"); 
    exit; 
} 

// If the user is logged in, we can safely proceed. 
$users = $SimpleUsers->getUsers(); 


if ($mysqli->connect_error) { 
die("Connection failed: ". $mysqli->connect_error); 
} 

if(isset($_POST['new']) && $_POST['new']==1) 
{ 

$aantaluren =$_REQUEST['aantaluren']; 
$datum =$_REQUEST['datum']; 
$projectname = $_REQUEST['projectname']; 

$sql = "INSERT INTO uren (aantaluren, projectname, datum) 
VALUES ('$aantaluren','$projectname','$datum'"; 

if ($mysqli->query($sql) === TRUE) { 
    echo "Uren succesvol toegevoegd. <a href='overzicht.php'> Bekijk  overzicht</a>"; 
} else { 
    echo "Error: " . $sql . "<br>" . $mysqli->error; 
}} 


?> 


<!DOCTYPE html> 
<html> 
<head> 
    <title>Toevoegen</title> 
    <link rel="stylesheet" href="style.css"> 
    </head> 
    <body> 
     <div class="form"> 
      <p> 
       <a href="dashboard.php">Dashboard</a> | <a href="view.php">Projecten inzien</a> | <a href="logout.php">Loguit</a> 
      </p> 

      <div> 
       <h1>Voeg uren toe</h1> 

       <form action="uren.php" method="post"> 
        <input type="hidden" name="new" value="1" /> 
        <p>Aantal uren</p><p> 
         <input name="aantaluren" type="number" min=1 max=24> 
         </p> 
         <select name="projectname"/> 

         <?php 
        $sql = mysqli_query($mysqli, "SELECT name FROM projecten"); 
        while ($row = $sql->fetch_assoc()){ 
         echo "<option value=\"dropdown\">" . $row['name'] . "</option>"; 
         } 
        ?> 

        </select> 
        <p>Datum</p> 
        <p> 
         <input type="date" name="datum" placeholder="datum" required /> 
        </p> 
        <p> 
         <input name="submit" type="submit" value="Voeg toe" /> 
        </p> 
       </form> 

      </div> 
     </div> 
    </body> 
</html>

这些都是我的SQL表：

CREATE DATABASE IF NOT EXISTS urenregistratie; 

CREATE TABLE IF NOT EXISTS urenregistratie.`users` (
    `userId` int(11) NOT NULL auto_increment, 
    `uUsername` varchar(128) NOT NULL, 
    `uPassword` varchar(40) NOT NULL, 
    `uSalt` varchar(128) NOT NULL, 
    `uActivity` datetime NOT NULL, 
    `uCreated` datetime NOT NULL, 
    PRIMARY KEY (`userId`), 
    UNIQUE KEY `uUsername` (`uUsername`) 
    ) ENGINE=MyISAM AUTO_INCREMENT=1 ; 

CREATE TABLE IF NOT EXISTS urenregistratie.`users_information` (
    `userId` int(11) NOT NULL, 
    `infoKey` varchar(128) NOT NULL, 
    `InfoValue` text NOT NULL, 
    KEY `userId` (`userId`) 
) ENGINE=MyISAM; 

CREATE TABLE IF NOT EXISTS urenregistratie.`projecten`(
    `projectId` int(11) NOT NULL AUTO_INCREMENT, 
    `trn_date` datetime NOT NULL, 
    `name` varchar(50) NOT NULL, 
    `begindatum` datetime NOT NULL, 
    `einddatum` datetime NOT NULL, 
    PRIMARY KEY (`projectId`) 
    ); 

CREATE TABLE IF NOT EXISTS urenregistratie.`uren`(
    projectname varchar(50), 
    aantaluren int(11) NOT NULL, 
    datum datetime NOT NULL, 
    PRIMARY KEY (projectname) 
    )

我一直停留在这大约2天了，我似乎无法弄清楚为什么它不起作用。请帮助

来源

2017-02-10 Erik Venema

您忘记关闭插入语句中的括号。完成。

正确的格式：

INSERT INTO uren（）VALUES（）

来源

2017-02-10 10:51:25 JordanH

谢谢这么多，它最终将它添加到数据库。现在我发现了另一个问题。我在那里有一个下拉菜单，但是当我选择一些内容时，它总是在下拉菜单中添加名称“Dropdown”而不是选定的选项。你知道这个问题是什么吗？ –

这听起来像您需要将下拉列表中的值更改为指定的字符串：例如。 value =“exampleString” – JordanH

@ErikVenema在'value'属性中给出行的值，或者移除'value'属性，因为显示的值将默认添加'echo“”;' – affaz

你没有关闭托架

$sql = "INSERT INTO uren (aantaluren, projectname, datum) 
VALUES ('$aantaluren','$projectname','$datum')";

它在你的错误说，它的语法错误

来源

2017-02-10 10:52:50 affaz

将数据插入SQL表时出错

回答

相关问题