Question

我需要比较表中的2个不同列：tbl_Try（country，name）。如果有2个具有相同值的coloumns什么都不做。如果这两个列的相同值没有相同的插入。

这是与数据库的连接：

      <?php
        require_once("menu.php");
        require_once("function.php");

    ?>

这是主要代码（阅读试图理解的评论）

            <?php
             $conn = ConnectToSql();

             $query= "Select *  FROM tbl_countries";
             $result = mysqli_query($conn, $query)
             or die("Error in query: ". mysqli_error($conn));

             $choose = '';

            while ($row = mysqli_fetch_assoc($result))
            {

                $choose .= '<option value = "'.$row['name'].'">'.$row['name'].'</option>';
            }

            ?>
        <div class="form-group">
          <label class="control-label col-sm-2" for="Country">Choose a country:</label>
          <div class="col-sm-5">
              <select class="form-control" name ="reg_country" >
                  <option></option>
                  <?php echo $choose;?></select>
          </div>
            </div>
            <br>
            <div class="form-group">
          <div class="col-sm-offset-4 col-sm-10">
            <button type="submit" name="submit" class="btn btn-default">Submit</button>
          </div>
        </div>

            <?php

             if(isset($_POST['submit']))
             {

                $country = $_POST['reg_country'];

                $_SESSION['country'] = $country;




                 $query2 = "SELECT name,id FROM tbl_flowertypes ";
                 $result2 = mysqli_query($conn, $query2) or die("Error in query: ". mysqli_error($conn));
                 $result2_rows = mysqli_num_rows($result2);


              //this code is not working. Need to count the 2 columns (name AND country) 
              // if in tbl_try there is already country:italy name:12 red roses. DO NOTHING if there is no data the same
              // insert data
                 $query3 = "SELECT count(*) FROM tbl_try WHERE name='$_SESSION[flower_type_name]' AND country ='$_SESSION[country]' ";


                 $result3 = mysqli_query($conn, $query3) or die("Error in query: ". mysqli_error($conn));
                 $result3_rows = mysqli_num_rows($result3);




               // loop counting how many record are in table flowertype and loop to insert data in tbl_try (this code is working fine)     
            for($i = 1; $i <= $result2_rows;$i++)
                {



                    while($row = mysqli_fetch_assoc($result2))
                        {

                            $_SESSION['flower_type_name'] = $row['name'];
                        // this is the insertion of the data
                            $inserting = "INSERT INTO tbl_try(name,country) VALUES ('$row[name]','$_SESSION[country]')";
                            $result3 = mysqli_query($conn, $inserting) or die("Error in query: ". mysqli_error($conn));

                        }



                }
             }

结果需要：

数据库需要插入一次。只要表中没有相同的栏目。如果有相同的数据，那就没有了。

我需要通过显示包含一些记录的表来完成此代码，但我知道如何做到这一点。

任何建议？

Answer 1

最好的方法是将这两列的唯一定义添加到DB中，如下所示：

ALTER TABLE tbl_Try
ADD UNIQUE (name, country)

然后你可以检查DB中的结果代码，并根据你现在的结果代码，如果因为大小写存储到DB，那么DB中已存在相同的名称和国家。

比较MySql中的多个coloumns

1 个答案: