Question

我正在构建一个网页搜索，该网页应列出不包含所列成分的食谱（用户可以输入一到四种成分，只需要第一种成分）。

我无法发布图片，因此我会尝试以文字显示。

搜索字段（用户可以输入一到四种成分，只需要第一种成分）

横幅：食谱缺少这些成分。

成分1的文本字段（必填）：香草

成分2的文本字段（可选）：牛肉

成分3的文本字段（可选）：tuna

成分4的文本字段（可选）：duck

搜索按钮

<?php

    //CONNECT TO DATABASE
    include_once ("dbc.php");

    //DECLAIR SESSION USERNAME
    $username=$_SESSION['username'];
    $getMemberID = mysql_query ("select member_id from members where member_username = '$username'");
    $memberID = mysql_result($getMemberID,0);

    //GET RECIPES WITHOUT INGREDIENT
    if (isset($_GET['ingredient1'])) {

        $ingredient1=$_GET['ingredient1'];
        //GET RECIPES FROM DATABASE
        $getAllRecipes = mysql_query ("select recipes.recipe_id, recipes.member_id, recipes.recipe_name as 'recipeName', cooking_time.cooking_time as 'cookingTime', concat(member_name, ' ', member_surname) as 'author' from members left join recipes ON members.member_id = recipes.member_id left join cooking_time ON recipes.cooking_time_id = cooking_time.cooking_time_id where recipes.recipe_ingredients not like '%$ingredient1%'");

    } else if (isset($_GET['ingredient1'], $_GET['ingredient2'])){

        $ingredient1=$_GET['ingredient1'];
        $ingredient2=$_GET['ingredient2'];
        //GET RECIPES FROM DATABASE
        $getAllRecipes = mysql_query ("select recipes.recipe_id, recipes.member_id, recipes.recipe_name as 'recipeName', cooking_time.cooking_time as 'cookingTime', concat(member_name, ' ', member_surname) as 'author' from members left join recipes ON members.member_id = recipes.member_id left join cooking_time ON recipes.cooking_time_id = cooking_time.cooking_time_id where recipes.recipe_ingredients not like '%$ingredient1%' and recipes.recipe_ingredients not like '%$ingredient2%'");

    } else if (isset($_GET['ingredient1'], $_GET['ingredient2'], $_GET['ingredient3'])) {

        $ingredient1=$_GET['ingredient1'];
        $ingredient2=$_GET['ingredient2'];
        $ingredient3=$_GET['ingredient3'];
        //GET RECIPES FROM DATABASE
        $getAllRecipes = mysql_query ("select recipes.recipe_id, recipes.member_id, recipes.recipe_name as 'recipeName', cooking_time.cooking_time as 'cookingTime', concat(member_name, ' ', member_surname) as 'author' from members left join recipes ON members.member_id = recipes.member_id left join cooking_time ON recipes.cooking_time_id = cooking_time.cooking_time_id where recipes.recipe_ingredients not like '%$ingredient1%' and recipes.recipe_ingredients not like '%$ingredient2%' and recipes.recipe_ingredients not like '%$ingredient3%'");

    } else if (isset($_GET['ingredient1'], $_GET['ingredient2'], $_GET['ingredient3'], $_GET['ingredient4'])) {

        $ingredient1=$_GET['ingredient1'];
        $ingredient2=$_GET['ingredient2'];
        $ingredient3=$_GET['ingredient3'];
        $ingredient4=$_GET['ingredient4'];
        //GET RECIPES FROM DATABASE
        $getAllRecipes = mysql_query ("select recipes.recipe_id, recipes.member_id, recipes.recipe_name as 'recipeName', cooking_time.cooking_time as 'cookingTime', concat(member_name, ' ', member_surname) as 'author' from members left join recipes ON members.member_id = recipes.member_id left join cooking_time ON recipes.cooking_time_id = cooking_time.cooking_time_id where recipes.recipe_ingredients not like '%$ingredient1%' and recipes.recipe_ingredients not like '%$ingredient2%' and recipes.recipe_ingredients not like '%$ingredient3%' and recipes.recipe_ingredients not like '%$ingredient4%'");

    } 

    //POPULATE TABLE ROWS WITH DATA FROM DATABASE
    while ($allRecipes = mysql_fetch_array ($getAllRecipes)) {

        //GET USERNAME TO USE FOR ENABLING MODIFY/DELETE    
        $dbMemberID = $allRecipes['member_id'];
        $getDbUsername = mysql_query ("select members.member_username from members where members.member_id = '$dbMemberID'");
        $dbUsername = mysql_result($getDbUsername,0);

        //CREATING TABLE ROWS WITH RECIPE INFORMATION
        echo "<tr>";
            echo "<td><a href='show_recipe.php?id=" . $allRecipes['recipe_id'] . "'>" . $allRecipes['recipeName'] . "<a></td>";         
            echo "<td>" . $allRecipes['cookingTime'] . "</td>";
            echo "<td>" . $allRecipes['author'] . "</td>";

            if ($username === $dbUsername) {
                echo "<td align='center'><a href='modify_recipe.php?id=" . $allRecipes['recipe_id'] . "'>Modify<a></td>";
                echo "<td align='center'><a href='delete_recipe.php?id=" . $allRecipes['recipe_id'] . "'>Delete<a></td>";
            } else {
                echo "<td></td>";
                echo "<td></td>";
            }

        echo "</tr>";

    }   
?>

现在的方式，即使我输入了四种成分（见下文），也只考虑输入的第一种成分。

网页中显示的搜索结果（我只显示食谱名称）：

食谱名称

Rib Roast

金枪鱼红薯夹克

通心粉奶酪配培根＆amp;松子

Duck confit burger

Chilli con carne

Lamb＆amp;生菜煎炸

番茄＆amp;洋葱沙拉

三重奶酪＆amp;茄子烤宽面条

煎三文鱼配豆瓣菜，玉米粥油炸面包块＆amp;刺山柑

燕麦华夫饼干

Kale和Portobello Lasagna

在上面，虽然我搜索不含香草，牛肉，金枪鱼和鸭肉的食谱，但我仍然得到含有牛肉，金枪鱼和鸭肉的食谱。我在工作台上尝试了同样的事情并获得了成功的结果：

    select 
    recipes.recipe_id,
    recipes.member_id,
    recipes.recipe_name as 'recipeName',
    cooking_time.cooking_time as 'cookingTime',
    concat(member_name, ' ', member_surname) as 'author'
from
    members
        left join
    recipes ON members.member_id = recipes.member_id
        left join
    cooking_time ON recipes.cooking_time_id = cooking_time.cooking_time_id
where
    recipes.recipe_ingredients not like '%Vanilla%'
        and recipes.recipe_ingredients not like '%beef%'
        and recipes.recipe_ingredients not like '%tuna%'
        and recipes.recipe_ingredients not like '%duck%';

工作台的结果（好）：

RecipeName中

通心粉奶酪配培根＆amp;松子

Lamb＆amp;生菜煎炸

番茄＆amp;洋葱沙拉

三重奶酪＆amp;茄子烤宽面条

煎三文鱼配豆瓣菜，玉米粥油炸面包块＆amp;酸豆

燕麦华夫饼干

Kale和Portobello Lasagna

任何人都可以帮助修改php代码吗？

提前致谢

Answer 1

你应该使用OR代替AND，因为它现在只会查找没有所有给定项目（香草，牛肉，金枪鱼，鸭子）的东西，只有一个，两个或其中三个会出现在你的搜索中。

请改为尝试：

select 
recipes.recipe_id,
recipes.member_id,
recipes.recipe_name as 'recipeName',
cooking_time.cooking_time as 'cookingTime',
concat(member_name, ' ', member_surname) as 'author'
from
members
    left join
recipes ON members.member_id = recipes.member_id
    left join
cooking_time ON recipes.cooking_time_id = cooking_time.cooking_time_id
where
recipes.recipe_ingredients not like '%Vanilla%'
    OR recipes.recipe_ingredients not like '%beef%'
    OR recipes.recipe_ingredients not like '%tuna%'
    OR recipes.recipe_ingredients not like '%duck%';

Answer 2

问题是检查的顺序，让我们说你得到所有4种成分。

首先检查isset($_GET['ingredient1']) 设置了 Ingedient1 ？当然！然后你的sql-query生成并执行。没有进一步的检查。

您应该将if语句从所有成分重建为一个：

if (isset($_GET['i1'], $_GET['i2'], $_GET['i3'], $_GET['i4'])) {
    // do something
} else if (isset($_GET['i1'], $_GET['i2'], $_GET['i3'])) {
    // do something
}  else if (isset($_GET['i1'], $_GET['i2'])) {
    // do something
} else if (isset($_GET['i1'])) {
    // do something
}

if isset php的问题

2 个答案: