Question

我的代码正在满足我的需求。但唯一让我烦恼的是＆＃34;其他＆＃34;不管用。当我搜索正确的记录时记录将出现，它运行正常。但如果我不正确搜索记录什么都不会发生。我期待＆＃34;未找到记录＆＃34;会回应但没有任何反应。

}else{
       echo "Records not found";
    }

这是整个代码。

      <?php
$conn = mysqli_connect("localhost", "root", "", "my1stdb") or die("could not connect");

$set = $_POST['search'];


if ($set) {
    $show   = "SELECT * FROM users where email='$set'";
    $result = mysqli_query($conn, $show);
    while ($rows = mysqli_fetch_array($result)) {
        echo "Registrant Found";
        echo "<tr>";
        echo "<td>";
        echo $rows['username'];
        echo "</td>";
        echo "<td>";
        echo $rows['fullname'];
        echo "</td>";
        echo "<td>";
        echo $rows['password'];
        echo "</td>";
        echo "<td>";
        echo $rows['email'];
        echo "</td>";
        echo "</tr>";
        echo "<br/>";
    }

} else {
    echo "Records not found";
}

?>
</table>

Answer 1

您需要使用mysqli_num_rows()和mysqli_fetch_assoc()： -

<?php
    $conn=mysqli_connect("localhost","root","","my1stdb") or die("could not connect");

    $set = $_POST['search']; 


    if($set) {
        $show="SELECT * FROM users where email='$set'";
        $result=mysqli_query($conn,$show) or die(mysqli_error($conn));
        if(mysqli_num_rows($result)>0){ // check data present or not
            while($rows=mysqli_fetch_assoc($result)){ // for lighter array due to associative indexes only
                echo "Registrant Found";
                echo "<tr>";
                echo "<td>";
                echo $rows['username'];
                echo "</td>";
                echo "<td>";
                echo $rows['fullname'];
                echo "</td>";
                echo "<td>";
                echo $rows['password'];
                echo "</td>";
                echo "<td>";
                echo $rows['email'];
                echo "</td>";
                echo "</tr>";
                echo "<br/>";
            }
        }else{
            echo "Records not found";
        }
    }else{
        echo "Please insert search term";
    }

?>
</table>

注意： - 您的代码是SQL INJECTION的开放式代码。防止使用prepared statements

<强> 参考： -

mysqli prepared statements

PDO prepared statements

Answer 2

您可以计算返回的结果数。

if($set) {
    $show="SELECT * FROM users where email='$set'";
    $result=mysqli_query($conn,$show);

    $recordCount = 0;

    while($rows=mysqli_fetch_array($result)){
        $recordCount++;
            echo "Registrant Found";
            echo "<tr>";
            echo "<td>";
            echo $rows['username'];
            echo "</td>";
            echo "<td>";
            echo $rows['fullname'];
            echo "</td>";
            echo "<td>";
            echo $rows['password'];
            echo "</td>";
            echo "<td>";
            echo $rows['email'];
            echo "</td>";
            echo "</tr>";
            echo "<br/>";
    }
    if($recordCount==0){
        echo "Records not found";
    }

}

简单的其他与循环不工作

2 个答案: