Question

我一直试图在我的日期数据库中添加信息，或者通过我的html页面将其删除。但是，当按下提交按钮时，什么也没有发生。 html页面将仅刷新而不对数据库进行任何更改。我不明白怎么了。
下面的代码显示了数据库以及用于发送信息

<?php
$tjener="localhost";
$brukernavn="root";
$passord="";
$database="treningsgruppe";

$kobling=new mysqli($tjener, $brukernavn, $passord, $database); 
if($kobling->connect_error) {
    die("Noe gikk galt: " . $kobling->connect_error);
} 
$kobling->set_charset("utf8");  
?> 
<?php 
$sql="SELECT * FROM treningsokter ORDER BY treningsokter_id DESC"; 
$resultat=$kobling->query($sql); 

echo "<table>"; 
echo "<tr>";
echo "<th>Treningsøkt_id</th>"; 
echo "<th>Løype_id</th>"; 
echo "<th>Person_id</th>";
echo "<th>Tid</th>";
echo "<th>Dato</th>";
echo "<th>Vær</th>"; 
echo "<th>Skade</th>"; 
echo "<th>Sykdom</th>"; 
echo "<th>Konkurranse?</th>"; 
echo "</tr>"; 

 while($rad=$resultat->fetch_assoc()) {
    $t_id=$rad["treningsokter_id"];
    $l_id=$rad["loype_id"];
    $p_id=$rad["person_id"]; 
    $tid=$rad["tid"];
    $dato=$rad["dato"];
    $vaer=$rad["vaer"]; 
    $skade=$rad["skade"];
    $sykdom=$rad["sykdom"];
    $konkurranse=$rad["konkurranse"];

    echo "<tr>";
    echo "<td>$t_id</td>";
    echo "<td>$l_id</td>";
    echo "<td>$p_id</td>";
    echo "<td>$tid</td>";
    echo "<td>$dato</td>";
    echo "<td>$vaer</td>";
    echo "<td>$skade</td>";
    echo "<td>$sykdom</td>";
    echo "<td>$konkurranse</td>";

    echo "<td><form method='POST'>";
    echo "<input type='hidden' name='slett_id' value='$t_id'>"; 
    echo "<input type='submit' name='slett' Value='Slett?'></form></td>";

    echo "</tr>"; 

}
echo "</table>";
?>

下面的代码用于INSERT INTO或DELETE FROM（下面的代码位于第一个代码之上）

<?php
$tjener="localhost";
$brukernavn="root";
$passord="";
$database="treningsgruppe";

$kobling=new mysqli($tjener, $brukernavn, $passord, $database); 

if($kobling->connect_error) {
    die("Noe gikk galt: " . $kobling->connect_error);
} 
$kobling->set_charset("utf8"); 

if(isset($_POST["leggtil"])) {
$l_id=$_POST["loype_id"]; 
$p_id=$_POST["person_id"]; 
$tid=$_POST["tid"];
$dato=$_POST["dato"];
$vaer=$_POST["vaer"];
$skade=$_POST["skade"];
$sykdom=$_POST["sykdom"];
$konkurranse=$_POST["konkurranse"];
$sql = "INSERT INTO treningsokter (loype_id, person_id, tid, dato, skade, sykdom, konkurranse) VALUES ('$l_id','$p_id','$tid','$dato','$vaer','$skade','$sykdom', '$konkurranse')";
$kobling->query($sql);
   }      
if(isset($_POST["slett"])) {
$slett_id = $_POST["slett_id"];
$sql = "DELETE FROM treningsokter WHERE treningsokt_id=$slett_id";
$kobling->query($sql);

} ？> 这是针对高中项目的。谢谢！

我收到此错误消息：

( ! ) Fatal error: Uncaught mysqli_sql_exception: Column count doesn't match value count at row 1 in C:\wamp64\www\Younes\Administrer_treningsokter.php on line 48

我不明白。第48行是INSERT INTO之后的$kobling->query($sql);。

在我的php代码中无法插入和删除

0 个答案: