基于完成的搜索搜索mysql数据库

时间:2012-03-16 01:27:38

标签: php mysql ajax arrays while-loop

我有3个链选择框,它从表中的同一行数据返回三个值:$ drop,$ drop_2和$ drop_3,我想运行一个mysql查询从该行的另一列返回一个值在表中。我可以直接将值“回显”到页面,但是我想使用数组值然后在“while”循环中从不同的表生成匹配结果。我有以下设置,如果有人能告诉我它为什么不起作用,我将非常感谢!

<h1>Search Results</h1>

<table>
<tr>
<th>Name</th>
<th>Number</th>
<th>Description</th>
</tr>

<?php

$search_results = mysql_query("SELECT id_sensor FROM vehicles_new WHERE make='$drop' AND model='$drop_2' AND year='$drop_3'");

while($result = mysql_fetch_array($search_results)){        

    echo $result['id_sensor']; //This is displaying the value I need to use in the next query
    ?>

<?php

$sql = mysql_query("SELECT * FROM products WHERE id='$result'");

    while($row = mysql_fetch_assoc($sql)){
?>
    <tr>
    <td><?php echo $row['id']; ?></td>
    <td><?php echo $row['part_id']; ?></td>
    <td><?php echo $row['description']; ?></td>
    </tr>
<?php
    }
?>

    <?php
    }
?>
</table>

提前谢谢,乔

1 个答案:

答案 0 :(得分:2)

更改此

$sql = mysql_query("SELECT * FROM products WHERE id='$result'");

到这个

$sql = mysql_query("SELECT * FROM products WHERE id='" . $result['id_sensor'] . "'");