Question

最后让它发挥作用后，我想我会发布它以防万一我在填充几组下拉选择标签时遇到了麻烦我研究了许多类似的提交和解决方案，但我仍然无法找到我一直在寻找的答案。

@Ronser帮助我测试了我的查询，这让我更多地了解了数组的实际工作方式。我意识到我需要返回并将表1 中的访问列更新为 access_id 。（我应该将这些原始索引）。

表1： app_generalData
APP_ID，
标题，
STATUS_ID，
CATEGORY_ID，
标签，
access_id

表2： app_access
access_id，
access_title

期望的结果：
目标1：显示/回显所选选项（存储在 app_access 表中）

目标2：
使用变量构建这些查询，以便轻松更新以添加新的下拉列表。

结果HTML：

<select name="access"><option "">Global</option>\n<option " selected ">Corporate</option>\n<option "">Local Site</option>\n</select>

代码：

<?php

//connect to the database
require_once('connectvars.php');

// global variable for this populating this dropdown
$dropdown =         "access";
$before_var =       "app_";
$column1=           $before_var.$dropdown;
$after_var =        "_title";
$column2=           $dropdown.$after_var;
$id_var=            "_id";
$dropdown_table_id= $dropdown.$id_var;

$optionsList = array();

$dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME)
    or die ('Error connecting to MySQL server.');

echo '<select name="' . $dropdown . '">';

// See if we're viewing a selected app or are we creating a new app
if (isset($_GET['app_id'])) {
          // print_r($_GET); // GET is Successful

  // 'if' [app_id] is appended in the url

      // STEP 1: Get the stored value of the SELECTED from mysql

          // Get "selected item" (id) from app_generalData (table) with the "selected app" (id)
          $option = "SELECT  ".$dropdown_table_id." FROM app_generalData WHERE app_id ='" . $_GET['app_id'] . "'";

          // submit the select statement
          // Get & store the value of "selected" <option>
          $selected_option = mysqli_query($dbc, $option) 
              or die(mysql_error());
          $row_1=mysqli_fetch_array($selected_option);

      // STEP 2: Build the SELECT input

          // Set the array of data to populate dropdown list <option>s
          $options = "SELECT * FROM ".$column1." ORDER BY ".$dropdown_table_id."";
                // NOTE: print_r($options)...SELECT access_title FROM app_access ORDER BY access_id
            $selected_options = mysqli_query($dbc, $options)
                or die(mysqli_error());

            $kk     = 0;   //initialize kk
           while($row_2 = mysqli_fetch_array($selected_options)) {

                $selected ='';
                if($row_1["$dropdown_table_id"]==$row_2["$dropdown_table_id"]) {
                $selected=' selected ';
                }
                $optionsList[$kk++] ='<option "' . $selected . '">' . $row_2["$column2"] . '</option>';
           }

          // Echo the <option>s
              $optionCount = count($optionsList);
              for($i=0;$i<$optionCount;$i++) {
                  echo $optionsList[$i].'\n';
              }

    }
    else {
        // Action 'if' no [app_id] is appended in the url
    };
    // close the last <select> tag
    echo '</select>';

    // close the last database
    mysqli_close($dbc); 

?>

Answer 1

请尝试以下方法......

检查＆＃39; app_id＆＃39;。
打印sql查询并直接在mysql中运行。

如果没有返回任何行或错误，请验证sql查询。

Answer 2

试试这个......

     $options = "SELECT ".$column2." FROM ".$column1." ORDER BY ".$dropdown_table_id."";
      $kk     = 0;   //initialize kk
     while($row = mysqli_fetch_array($options)) {
          $selected ='';
          if($selected_option==$row["$column2"]) {
          $selected=' selected ';
      }
      $optionsList[$kk++] ='<option "' . $selected . '">' . $row["$column2"] . '</option>';          
     //try this change. or

     echo '<option "' . $selected . '">' . $row["$column2"] . '</option>'; //print it here itself
     }

    print_r($optionsList);

Answer 3

<?php

//connect to the database
require_once('connectvars.php');

// global variable for this populating this dropdown
$dropdown =         "access";
$before_var =       "app_";
$column1=           $before_var.$dropdown;
$after_var =        "_title";
$column2=           $dropdown.$after_var;
$id_var=            "_id";
$dropdown_table_id= $dropdown.$id_var;

$optionsList = array();

$dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME)
    or die ('Error connecting to MySQL server.');

echo '<select name="' . $dropdown . '">';

// See if we're viewing a selected app or are we creating a new app
if (isset($_GET['app_id'])) {
          // print_r($_GET); // GET is Successful

  // 'if' [app_id] is appended in the url

      // STEP 1: Get the stored value of the SELECTED from mysql

          // Get "selected item" (id) from app_generalData (table) with the "selected app" (id)
          $option = "SELECT  ".$dropdown_table_id." FROM app_generalData WHERE app_id ='" . $_GET['app_id'] . "'";

          // submit the select statement
          // Get & store the value of "selected" <option>
          $selected_option = mysqli_query($dbc, $option) 
              or die(mysql_error());
          $row_1=mysqli_fetch_array($selected_option);

      // STEP 2: Build the SELECT "selected" and <options>

          // Set the array of data to populate dropdown list <option>s
          $options = "SELECT * FROM ".$column1." ORDER BY ".$dropdown_table_id."";
                // NOTE: print_r($options)...SELECT access_title FROM app_access ORDER BY access_id
            $selected_options = mysqli_query($dbc, $options)
                or die(mysqli_error());

            $kk     = 0;   //initialize kk
           while($row_2 = mysqli_fetch_array($selected_options)) {

                $selected ='';
                // Compare access_id from table1 against access_id from table2
                if($row_1["$dropdown_table_id"]==$row_2["$dropdown_table_id"]) {
                $selected=' selected ';
                }
                $optionsList[$kk++] ='<option "' . $selected . '">' . $row_2["$column2"] . '</option>';
           }

          // Echo the <option>s
              $optionCount = count($optionsList);
              for($i=0;$i<$optionCount;$i++) {
                  echo $optionsList[$i].'\n';
              }

    }
    else {
        // Action 'if' no [app_id] is appended in the url
    };
    // close the last <select> tag
    echo '</select>';

    // close the last database
    mysqli_close($dbc); 

?>

使用mysql中的变量动态选择选项

3 个答案: