Question

我创建了我的个人资料页面，显示来自MySQL的所有表格数据。所有数据都在表单和下拉列表中正确显示。但问题是选项值在选项列表中显示两次。

这是我的代码：

DEBUG JdbcTemplate - Executing prepared SQL statement [Select Account_ID,password,activated from Account where email=?]
DEBUG DataSourceUtils - Fetching JDBC Connection from DataSource
DEBUG DriverManagerDataSource - Creating new JDBC DriverManager Connection to [jdbc:mysql://db4free.net:3306/frozencat?useSSL=false]
TRACE StatementCreatorUtils - Setting SQL statement parameter value: column index 1, parameter value [admin@admin.de], value class [java.lang.String], SQL type unknown
DEBUG DataSourceUtils - Returning JDBC Connection to DataSource
DEBUG JdbcTemplate - Executing prepared SQL query
DEBUG JdbcTemplate - Executing prepared SQL statement [SELECT Account_ID,role FROM Account_Roles WHERE Account_ID=?]
DEBUG DataSourceUtils - Fetching JDBC Connection from DataSource
DEBUG DriverManagerDataSource - Creating new JDBC DriverManager Connection to [jdbc:mysql://db4free.net:3306/frozencat?useSSL=false]
TRACE StatementCreatorUtils - Setting SQL statement parameter value: column index 1, parameter value [1], value class [java.lang.String], SQL type unknown
DEBUG DataSourceUtils - Returning JDBC Connection to DataSource
INFO  AuthenticationEventListener - Login attempt with username: 1      Success: true

Answer 1

您需要更改以下while代码： -

while($country = mysqli_fetch_array($result)){
    //Compare User Country with country list. $row[4] is the country column in user table
    if($row[4] == $country[1]){
        echo "<option value='$country[1]' selected='selected'>$country[1]</option>";
    }else{
        echo "<option value='$country[1]'>$country[1]</option>";
    }
}

注意：在您的代码中创建第一个选项，然后检查条件，这就是为什么它会显示所选选项的两倍。

Answer 2

您面临的问题的正确答案由 A-2-A 提供。

<强>另外您正在将所有循环选项嵌套在“＃34;选择国家/地区”中。选项。您应该删除</option>标记前的最后一个</select>标记，然后在＆＃34;选择国家/地区＆＃34;之后移动它像这样：

<option value="">Select Country</option>

Answer 3

这可以解决您的问题：

<select class="form-control" name="country" id="country">
 <option value="">Select Country
     <?php
        //Get country list from Country master
        $qry = "select * from country_master";
        //Execute query
        $result = mysqli_query($conn, $qry);
        //Assigned fetched array to $Country
        while($country = mysqli_fetch_array($result))
        {
             echo "<option value='$country[1]'".($row[4] == $country[1] ? " selected" : "").">$country[1]</option>";
        }
     ?>
  </option>

问题在于您已将选定选项另外回显到未选择的选项。现在，它添加了选择的＆＃39;如果应该选择该选项，则属性。（根据您的情况）

Answer 4

它应该像这样设置：

<select class="form-control" name="country" id="country">
     <option value="">Select Country
         <?php
            //Get country list from Country master
            $qry = "select * from country_master";
            //Execute query
            $result = mysqli_query($conn, $qry);
            //Assigned fetched array to $Country
            while($country = mysqli_fetch_array($result))
            {

              //Compare User Country with country list. $row[4] is the country column in user table
              if($row[4] == $country[1]){
                echo "<option value='$country[1]' selected='selected'>$country[1]</option>";
              }
              else{
                echo "<option value='$country[1]'>$country[1]</option>";
              }

            }
         ?>
      </option>
</select>

因为你需要检查是否选择了值，如果它没有相应地显示数据。

Answer 5

您有两个问题：

在选项中使用选项。
秒，您只需要打印所选属性而不是打印完整选项。

示例：

<option value="">Select Country </option> <?php //Get country list from Country master $qry = "select * from country_master"; //Execute query $result = mysqli_query($conn, $qry); //Assigned fetched array to $Country while($country = mysqli_fetch_array($result)) { if($row[4] == $country[1]) { $selected = 'selected=""'; } else{ $selected = ""; } ?> <option <?php echo $selected;?> value='<?php echo $country[1];?>'> <?php echo $country[1];?> </option> <?php } ?>

PHP＆amp; MySQLi - 在下拉列表中两次显示所选值

5 个答案: