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while 的变量传出去不一样?该如何解决

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Release: 2016-06-13 11:56:19
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while 的变量传出去不一样?

<br />function type_son_id_finder($type_id){<br />        $query=mysql_query("SELECT id FROM `protduct_type` WHERE `f_id` = '$type_id'");<br />        while ($row=mysql_fetch_assoc($query)) {<br />                return $all_son = $row["id"].",";<br />        }<br />}<br />
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这个方法是输入一个父栏目ID 就查找到有什么子栏目ID

以上的方法

应该是输出

1,2,

但只能输出
1,

但如果把
                return $all_son = $row["id"].",";
改为:
                echo $all_son = $row["id"].",";

就能正常的显示出1,2,

这是为什么?

我如何解决?
我是想把这个方法最终输出成

1,2

让我在mysql中 WHERE id IN (1,2) 

但现在只能输出第一个,还不能用return, 只能用echo?

求解!
------解决方案--------------------

function type_son_id_finder($type_id){
        $query=mysql_query("SELECT id FROM `protduct_type` WHERE `f_id` = '$type_id'");
        $all_son = '';
        while ($row=mysql_fetch_assoc($query)) {
                $all_son .= $row["id"].",";
        }
        return $all_son;
}

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