Question

我编写了这段代码，点击“提交”按钮后，它不会显示包含数据的表格。我检查过nazwa1是否包含一些字符串，但它不会显示任何内容。我是PHP和HTML的新手，所以我不确定这是好的代码。我读了一些网站，我认为我的代码看起来不错。我到处寻找答案，但我一无所获。谢谢你的帮助。

        <?php

                $link=mysqli_connect(connect_data);

                    if(mysqli_connect_errno()){
                    printf("Nie udało się połączyć: %s", mysqli_connect_error());
                    exit();};

        echo "<div class=\"row\">";
            $query = "SELECT Distinct Nazwa FROM vDane ";
            $result = mysqli_query($link, $query);
                echo "<div class=\"col-xl-12 marginesy\">";
                echo "<div class=\"row\">";
                    echo "<div class=\"col-xl-3\">";
                        echo "<select name=\"nazwa1\" class=\"form-control form-control-sm\"  placeholder=\"nazwa1\">";
                        echo "<option value=\"Pracownik\">Pracownik</option>";
                        while( $row = mysqli_fetch_array($result)){
                            echo "<option value=".$row['Nazwa'].">" .$row['Nazwa']. "</option>";

                        }
                        echo "</select>";
                    echo "</div>";
            $query = "SELECT Distinct Date FROM vDane ";
            $result = mysqli_query($link, $query);
                    echo "<div class=\"col-xl-3\">";
                        echo "<select class=\"form-control form-control-sm\" name=\"Date\" placeholder=\"Date\">";
                        echo "<option value=\"Data\">Data</option>";
                        while( $row = mysqli_fetch_array($result)){
                            echo "<option value=".$row['Date'].">" .$row['Date']. "</option>";

                        }                       
                        echo "</select>";
                        //$first_option = $_POST(['Date']);

                    echo "</div>";
                    echo "<div class=\"col-xl-3\">";
                        echo "<select class=\"form-control form-control-sm\" name=\"data\" placeholder=\"Data\">";
                        echo "</select>";
                    echo "</div>";
                    echo "<div class=\"col-xl-3\">";
                        echo "<form action=\"\" method=\"POST\">";
                        echo "<input class=\"btn btn-primary btn-sm \" type=\"submit\" name=\"find\" value=\"Submit\">";

                        echo "</select>";
                    echo "</div>";


                echo "</div>";
                echo "<div class=\"row tabela\"  style=\"overflow: scroll !important\">";           
                echo "<table class=\"table table-bordered table-md\">";
                echo "<thead>";
                    echo "<tr>";
                        echo "<th>Imię Nazwisko</th>";
                        echo "<th>Data</th>";
                        echo "<th>Godzina</th>";
                        echo "<th>Wej/wyj</th>";            
                    echo "</tr>";
                echo "</thead>";
                echo "<tbody>";
                if(isset($_POST['find'])){

                    $second_option = $_POST['nazwa1'];
                    echo "<div class=\"alert alert-success\" role=\"alert\">";
                        echo "<a href=\"#\" class=\"alert-link\">".$second_option."</a>";
                        echo "</div>";
                    if($second_option == "Pracownik"){
                        $query = "SELECT * FROM vDane ";
                        $result = mysqli_query($link, $query);
                        $row = mysqli_fetch_array($result);

                    }
                    if($second_option != "Pracownik"){
                        $query = "SELECT * FROM vDane WHERE Nazwa = $second_option";
                        $result = mysqli_query($link, $query);
                        $row = mysqli_fetch_array($result);         
                    }

                        foreach($row as $row){
                            echo "<tr>";
                            echo "<th scope=\"row\">".$row['Nazwa']."</th>";
                            echo "<td>".$row['Date']."</td>";
                            echo "<td>".$row['Time']."</td>";
                            echo "<td>".$row['Scan_ID']."</td>";
                            echo "</tr>";       
                        }
                    echo "</tbody>";
                    echo "</table>";
                    echo "</form>";



                    }
                mysqli_close($link);

                ?>

Answer 1

尝试在表名周围加上方括号或反引号。

COPY FROM

在我看来，它试图选择“Distinct”作为你的桌子，而不是“Distinct Nazwa”。根据一些Stack Overflow问题，如果这是问题，上面的一个应该解决它。

PHP表不显示数据

1 个答案: