需要帮助重构我的冗余代码（使用MySQL/PHP填充/构建3个选择列表）

我有一个膳食/食谱数据库，用于创建每日膳食计划。我需要创建3个不同的选择列表（早餐、午餐、晚餐），以显示每个餐名的可用食谱选项

到目前为止，我对上述每一项都使用单独的查询，并使用单独的构建来显示每个列表的结果

午餐查询：

  // Lunch options
  $sql = "SELECT plan_date,
    plan_meal,
    plan_recipe,
    recipe_name,
    recipe_serving_size
    FROM recipe_plans
    LEFT JOIN $table_meals ON meal_id = plan_meal    
    LEFT JOIN $table_recipes ON recipe_id = plan_recipe
    WHERE plan_date = '".$date."'
    AND meal_name = 'Lunch'
    AND plan_owner = '".$user_name."'
    ORDER BY recipe_name";
    $rc = $DB_LINK->Execute($sql);
    DBUtils::checkResult($rc, NULL, NULL, $sql);

  // Scan the rows of the SQL result
  while (!$rc->EOF) {
    $recipeList[($rc->fields['plan_recipe'])] = $rc->fields['recipe_name'] . " (" . $rc->fields['recipe_serving_size'] . ")";
    $rc->MoveNext();
  }

// Scan the fields in the SQL result row
// Print all existing Meals, and some new ones
$minShow = 1;
$maxShow = 1;  // only build 1 while testing
for ($i = 0; $i < (isset($MealPlanObj->mealplanItems[$date]) && ($i < $maxShow) ? count($MealPlanObj->mealplanItems[$date]) : 0) + $minShow; $i++) {
  if ($i < (isset($MealPlanObj->mealplanItems[$date]) ? count($MealPlanObj->mealplanItems[$date]) : 0)) {
    // If it is an existing meal item, then set it
    $meal = $MealPlanObj->mealplanItems[$date][$i]['meal']; // meal_id
    $servings = $MealPlanObj->mealplanItems[$date][$i]['servings'];
    $recipe = $MealPlanObj->mealplanItems[$date][$i]['id']; // recipe_id
  } else {
    // It is a new one, give it blank values
    $meal = NULL;
    $servings = $defaultServings;
    $recipe = NULL;
  }
  // The HTML Code to build the select list for 'Lunch'
}

和构建：

  // Lunch options
  $sql = "SELECT plan_date,
    plan_meal,
    plan_recipe,
    recipe_name,
    recipe_serving_size
    FROM recipe_plans
    LEFT JOIN $table_meals ON meal_id = plan_meal    
    LEFT JOIN $table_recipes ON recipe_id = plan_recipe
    WHERE plan_date = '".$date."'
    AND meal_name = 'Lunch'
    AND plan_owner = '".$user_name."'
    ORDER BY recipe_name";
    $rc = $DB_LINK->Execute($sql);
    DBUtils::checkResult($rc, NULL, NULL, $sql);

  // Scan the rows of the SQL result
  while (!$rc->EOF) {
    $recipeList[($rc->fields['plan_recipe'])] = $rc->fields['recipe_name'] . " (" . $rc->fields['recipe_serving_size'] . ")";
    $rc->MoveNext();
  }

// Scan the fields in the SQL result row
// Print all existing Meals, and some new ones
$minShow = 1;
$maxShow = 1;  // only build 1 while testing
for ($i = 0; $i < (isset($MealPlanObj->mealplanItems[$date]) && ($i < $maxShow) ? count($MealPlanObj->mealplanItems[$date]) : 0) + $minShow; $i++) {
  if ($i < (isset($MealPlanObj->mealplanItems[$date]) ? count($MealPlanObj->mealplanItems[$date]) : 0)) {
    // If it is an existing meal item, then set it
    $meal = $MealPlanObj->mealplanItems[$date][$i]['meal']; // meal_id
    $servings = $MealPlanObj->mealplanItems[$date][$i]['servings'];
    $recipe = $MealPlanObj->mealplanItems[$date][$i]['id']; // recipe_id
  } else {
    // It is a new one, give it blank values
    $meal = NULL;
    $servings = $defaultServings;
    $recipe = NULL;
  }
  // The HTML Code to build the select list for 'Lunch'
}

//扫描SQL结果行中的字段
//打印所有现有膳食和一些新膳食
$minShow=1；
$maxShow=1；//测试时仅构建1
对于（$i=0；$i<（isset（$MealPlanObj->mealplanItems[$date]）&（$i<$maxShow）？计数（$MealPlanObj->mealplanItems[$date]）：0）+$minShow；$i++）{
如果（$i<（设置（$MealPlanObj->mealplanItems[$date]）？计数（$MealPlanObj->mealplanItems[$date]）：0）{
//如果它是现有的膳食项目，则设置它
$MEIN=$MealPlanObj->mealplanItems[$date][$i]['MEIN']]；//膳食id
$servings=$MealPlanObj->mealplanItems[$date][$i]['servings']；
$recipe=$MealPlanObj->mealplanItems[$date][$i]['id']；//recipe\u id
}否则{
//这是一个新的，给它空白值
$MEIN=NULL；
$servings=$defaultServings；
$recipe=NULL；
}
//用于构建“午餐”选择列表的HTML代码
}

上面的代码对每顿饭的名字都是重复的，因为这就是我有限的技能留给我的地方，哈哈

---

问题是：与其为3个条件（早餐、午餐、晚餐）中的每一个编写单独的select和build语句，我怎样才能只编写1，将它们全部输出？

您已经使用了变量来构建SQL查询，因此您可以只引入另一个变量，如

$MEINE\u name

。此变量可应用于SQL语句。而不是：

$sql = "SELECT mplan_date,
  ...
  AND meal_name = 'Lunch'
  ...

然后你会写：

$meal_name = 'Lunch';
...
$sql = "SELECT mplan_date,
  ...
  AND meal_name = '" . mysqli_real_escape_string($meal_name) . "'
  ...

请注意函数

mysqli\u real\u escape\u string（）

的使用，尽管在本例中并非绝对必要，但如果您不能绝对确定变量内部的内容，则必须对添加到SQL语句中的所有变量进行转义。您的示例代码易受SQL注入攻击

之后，您可以进一步将代码打包到函数中：

function buildSqlQuery($meal_name, $date, $user_name)
{
  $sql = "SELECT mplan_date,
    ...
    ORDER BY recipe_name";
  return $sql;
}

$sqlForBreakfast = buildSqlQuery('Breakfast', '2000-01-01', 'teddy');
$sqlForLunch = buildSqlQuery('Lunch', '2000-01-01', 'teddy');
$sqlForDinner = buildSqlQuery('Dinner', '2000-01-01', 'teddy');

---

我不明白。php代码中没有对餐名的引用。问题是什么？此外，我认为您可能可以制作一些循环

foreach

，以使代码更易于阅读。因此，这有助于我解决问题的前半部分……但我如何为3个列表中的每一个实现实际构建的自动化？同样，我们将用于构建sql select语句的代码放入函数中，我们可以将执行查询和写入HTML的代码放入函数中。只需将变量

$sqlforfast

传递给查询函数，然后将

$recipeList

传递给编写函数。这三种情况我们都可以重复。^^^^这应该是显而易见的。我想我是想得太多而导致“分析瘫痪”，哈哈，又是thx！