之前代码验证mysql连接无误.
这里我想实现的是:activite返回1个以上的参数的话,对每个参数进行sql查询,然后把每个查询结果显示出来,可是我的是个死循环,可且就是返回一个参数,也不显示结果,请问是这么回事呢?
mysql_select_db($mysql_database,$link);
if (isset($_POST["activite"])) {
for($i=0;$i $strsql="select n_naps from activage where famille='$activite[$i]'";
echo $strsql;
$result=mysql_query($strsql);
$nb=mysql_num_rows($result);
for($i=1;$i $info=mysql_fetch_row($result);
echo "$info[0]
";
}
}
}
谢谢各位大神的指教.
回复讨论(解决方案)
出错了!
1、for($i=0;$i 没有看到对 $activite 赋值
2、$strsql="select n_naps from activage where famille='$activite[$i]'"
应写作
$strsql="select n_naps from activage where famille=' {$activite[$i] }'"
出错了!
1、for($i=0;$i 没有看到对 $activite 赋值
2、$strsql="select n_naps from activage where famille='$activite[$i]'"
应写作
$strsql="select n_naps from activage where f……
版主大人,谢谢你每次都有回答我的一些白痴问题啊,我把问题说的在细点吧:
在1.php中有代码如下:
Activités gymniques
Activités gymniques aquatiques
Activités fitness/forme
Activités douces/ bien-être
2.php接收1.php中的参数:局部代码如下:
之前验证连接数据库无误。
$activite=isset($_POST["activite"])?$_POST["activite"]:"";
if (isset($_POST["activite"])) {
for($i=0;$i $strsql="select n_naps from activage where famille='{$activite[$i]}'" ;
echo $strsql;
$result=mysql_query($strsql);
$nb=mysql_num_rows($result);
echo $nb;
for($i=1;$i $info=mysql_fetch_row($result);
echo "$info[0]
";
}
}
}
按照版主大人说的改了下,其中在1.php中单选了Activités gymniques 选项,上面代码结果输出为:
select n_naps from activage where famille='activités gymniques'0
不知道为什么查询结果是0,我复制粘贴的这段话放到mysql中,找到了4个结果.
这里是0,不知道为啥...
引用 1 楼 xuzuning 的回复:出错了!
1、for($i=0;$i 没有看到对 $activite 赋值
2、$strsql="select n_naps from activage where famille='$activite[$i]'"
应写作
$strsql="select n_naps f……
对2楼的补充啊,如果把where之后的都去掉,就显示出所有的n_naps了,就是说问题出在where哪里了,但是我实在找不出哪里有错啊....就是写$strsql="select n_naps from activage where famille='activités gymniques'还是说是0的......
字符 é 在中文操作系统中是2字节的,其内码与所用字符集有关

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