Question

我应该在此查询中输入if？我希望如果查询返回为空，则表单不会显示，但只显示类似于＆＃34的消息;表格中有更正的信息＆＃34; php很难做到这一点吗？

这是代码：

    //Step 2: Kjør en SQL-query to databasen 
$sql  = "SELECT * FROM oppgave  WHERE modulid=4 AND resultat is NULL ORDER BY RAND() LIMIT 1";

$data = null;
$dataid = null;
$result = mysql_query($sql, $tilkobling); 
echo "<hr>";
while ($nextrow= mysql_fetch_assoc($result)){
echo "Besvarelse: " . $nextrow['besvarelse'];
echo "<br>Modulid: " . $nextrow['modulid'];
//echo "<br>student: " . $nextrow['studentid'];
echo "<br>";
     $data = $nextrow['modulid'];
     $dataid = $nextrow['id'];
}



//Skjema til utfylling mot database
echo '<form name="input" action="tilretting.php" method="post">';
    echo'Retter<input type="text" name="correctedby" value="'.$_SESSION['myusername'].'">';
    echo '<input type="hidden" name="resultat" value="0">';
    echo 'Godkjent<input type="checkbox" name="resultat" value="1">';
    echo 'modul<input type="text" name="modulid" value="'.$data.'">';
     echo 'id<input type="text" name="id" value="'.$dataid.'">';
    echo '<input class="levermodulknapp" type="submit" name="lever1" value="Send inn retting">';
    echo  "</form>";
    echo "<hr>";
?>

Answer 1

使用empty功能

空 - 确定变量是否为空

if(!empty($nextrow)){

echo 'Show your Form';

}else{

echo 'HIde your Form';

}

Answer 2

在您的情况下，您应该只能查看$data和$dataid变量，例如

    if($data != null && $dataid != null){
        echo '<form name="input" action="tilretting.php" method="post">';
        echo'Retter<input type="text" name="correctedby" value="'.$_SESSION['myusername'].'">';
        echo '<input type="hidden" name="resultat" value="0">';
        echo 'Godkjent<input type="checkbox" name="resultat" value="1">';
        echo 'modul<input type="text" name="modulid" value="'.$data.'">';
         echo 'id<input type="text" name="id" value="'.$dataid.'">';
        echo '<input class="levermodulknapp" type="submit" name="lever1" value="Send inn retting">';
        echo  "</form>";
        echo "<hr>";
    }else{
        echo "there is noting to correct in the table";
    }

在查询中放置IF（MYSQL）

2 个答案: