Question

我希望这是一个简单的问题，我找不到任何完整问题的答案。

我有一个从MySQL返回的数组，看起来像这样

mysql> select c.comment, d.name, m.ounces from drink d 
       inner join mix m on m.drinkID = d.ID
       inner join cocktail c on m.cocktailID = c.id
       where c.id =5;
+----------------------+------------+--------+
| comment              | name       | ounces |
+----------------------+------------+--------+
| Long Island Iced Tea | Vodka      |   0.75 |
| Long Island Iced Tea | Gin        |   0.75 |
| Long Island Iced Tea | Rum        |   0.75 |
| Long Island Iced Tea | Tequila    |   0.75 |
| Long Island Iced Tea | Sour Mix   |   2.00 |
| Long Island Iced Tea | Cola       |   1.00 |
| Long Island Iced Tea | Triple Sec |   1.00 |
+----------------------+------------+--------+

我正在尝试让PHP生成一个类似于下面的无序列表。

<div class="jumbotron">
    <h2>
      Long Island
    </h2>
    <p>
      <ul>
        <li>0.5 oz Vodka</li>
        <li>0.5 oz Rum</li>
        <li>0.5 oz Gin</li>
        <li>0.5 oz Tequila</li>
        <li>0.5 oz Triple Sec</li>
        <li>1 oz Sweet and Sour Mix</li>
        <li>1 oz cola </li>
      </ul>

我正在尝试这个：

    $ingredient = "select 
       c.comment, d.name, m.ounces 
       from drink d 
       inner join mix m on m.drinkID = d.ID
       inner join cocktail c on m.cocktailID = c.id 
       where c.id = 4";
    //print($ingredient);
    $result = mysql_query($ingredient);

<?php foreach($result as $results): ?>
    <li><? echo $results ?></li>
<?php endforeach; ?>

从这里：StackOverflow但无法弄清楚为什么我一直从PHP接收HTTP 500错误。

任何帮助或指导都将不胜感激！

谢谢！

Answer 1

您正在使用最有可能被禁用的短标签。试试这个

WHERE

Answer 2

{"html_attributions": [], "result": { "address_components": [ { "long_name": "1", "short_name": "1", "types": [ "street_number" ] }, { "long_name": "East 161st Street", "short_name": "E 161st St", "types": [ "route" ] }, { "long_name": "Concourse", "short_name": "Concourse", "types": [ "neighborhood", "political" ] }, { "long_name": "Bronx", "short_name": "Bronx", "types": [ "sublocality_level_1", "sublocality", "political" ] }, { "long_name": "Bronx County", "short_name": "Bronx County", "types": [ "administrative_area_level_2", "political" ] },返回一组行，而不是单个行

mysql_query()

PHP数组到无序列表

2 个答案: