我想将数据从一种表格发送到许多表中，但是我选择的表却没有其他

Question

I am new to in programming. I want to send data into multiple tables by using one FORM but not every table just my selected table will be updated with data. It's similar to have an option when I select an option it will send only this table not others, I don't know is it a right method or wrong please help me...

    <div class="row">
        <div class="data-store text-center">
            <?php
                add_part('admin-menu.php');
            ?>
            <?php
                if(!empty($_POST)){
                    $title=$_POST['title'];
                    $sel_data='sel_data';
                    if($sel_data == 1){
                        $insert="INSERT INTO `database`.`gfx` (`gfx_id`, `gfx_title`) VALUES (NULL, '$title');";
                        if(mysqli_query($con,$insert)){
                            echo "Product has been uploaded";
                        }

                    }else{
                        echo "Failed to store data";
                    }

                    if($sel_data == 2){
                        $insert="INSERT INTO `database`.`gfx` (`gfx_id`, `gfx_title`) VALUES (NULL, '$title');";
                        if(mysqli_query($con,$insert)){
                            echo "Product has been uploaded";
                        }

                    }else{
                        echo "Failed to store data";
                    }



                }
            ?>

            <form action="" method="post" enctype="multipart/form-data">
                <div>
                    <label for="title"></label>
                    <input name="title" id="title" type="text" placeholder="post title"/> 
                </div>

                <select name="sel_data" id="">
                    <option name="" >Select one</option>
                    <option name="table1" value="1">Option 1</option>
                    <option name="table2" value="2">Option 2</option>
                    <option name="table3" value="3">Option 3</option>
                </select>

                <input type="submit" value="submit" />
            </form>
        </div>
    </div>

Answer 1

Select tables from your database using

" SELECT table_name FROM information_schema.tables WHERE table_schema = 'your_database_name' "

use resulted array for generating options using foreach. then instead of giving integer value for options, you should give name of table as value. after than you can change code as

        <?php
            if(!empty($_POST)){
                $title = $_POST['title'];
                $sel_data = $_POST['sel_data'];
                $insert="INSERT INTO `database`.`.$sel_data.` (`gfx_id`, `gfx_title`) VALUES (NULL, '$title');";
                if(mysqli_query($con,$insert)){
                  echo "Product has been uploaded";
                }else{
                  echo "Failed to store data";
                }
             }
        ?>

By using above code entered title will be inserted into selected table only.

Answer 2

You can optimize your code like this if all the table having the same structure.

<?php
if (!empty($_POST)) {
    $title = $_POST['title'];
    $sel_data = $_POST['sel_data'];
    if (!empty($sel_data)) {
        $insert = "INSERT INTO $sel_data.`gfx` (`gfx_id`, `gfx_title`) VALUES (NULL, '$title');";
        if (mysqli_query($con, $insert)) {
            echo "Product has been uploaded";
        }
    } else {
        echo "Failed to store data";
    }
}
?>

<div class="row">
    <div class="data-store text-center">
<?php
add_part('admin-menu.php');
?>           

        <form action="" method="post" enctype="multipart/form-data">
            <div>
                <label for="title"></label>
                <input name="title" id="title" type="text" placeholder="post title"/> 
            </div>

            <select name="sel_data" id="">
                <option  value="" >Select one</option>
                <option  value="table1">Option 1</option>
                <option  value="table2">Option 2</option>
                <option  value="table3">Option 3</option>
            </select>

            <input type="submit" value="submit" />
        </form>
    </div>
</div>

Answer 3

使用$ sel_data = $ _ POST ['sel_data']代替$ sel_data ='sel_data';

Answer 4

EDITED
hey have you tried elseif? and also i think the only wrong thing was your structure

    <?php
                    if(!empty($_POST)){
                        $title=$_POST['title'];
                        $sel_data=$_POST['sel_data']; // you need to call the value of the select
                        if($sel_data == 1){
                            $insert="INSERT INTO `database`.`gfx` (`gfx_id`, `gfx_title`) VALUES (NULL, '$title');";
                              if(mysqli_query($con,$insert)){
                                echo "Product has been uploaded";
                                }
                                else{
                                  echo "Failed to store data";
                            }
                        }


                        elseif($sel_data == 2){
                            $insert="INSERT INTO `database`.`gfx` (`gfx_id`, `gfx_title`) VALUES (NULL, '$title');";   
                            if(mysqli_query($con,$insert)){
                                echo "Product has been uploaded";
                            }
                            else{
                            echo "Failed to store data";
                              }
                            }

                         elseif($sel_data == 3){
                            $insert="INSERT INTO `database`.`gfx` (`gfx_id`, `gfx_title`) VALUES (NULL, '$title');";   
                             if(mysqli_query($con,$insert)){
                                echo "Product has been uploaded";
                            }
                            else{
                            echo "Failed to store data";
                                }
                            }

                        else {
                            echo"failed";
                        }
                    }