最简单的查询结果语句怎么写
我不会PHP,但是现在只有PHP的空间,
想要一个功能,很简单的一个,
数据表里就两列(A & B)
实现:我在文本框输入一个值,在数据库查询A列中这个值所对应的B列的值并返回给页面。
额...有大神能帮帮忙写个现成的较完整语句呗? 包括数据库连接语句部分的
回复讨论(解决方案)
error_reporting(0);
$conn=mysql_connect("localhost","root","root");//填写数据库连接信息
mysql_select_db("test");//填写数据库名
$valA = $_POST["valA"];
if($valA!=""){
$sql = "select * from test where a=".$valA;
$result=mysql_query($sql);
while($arr=mysql_fetch_array($result))
{
echo $arr["b"]."
";
}
}else{
?>
}
?>
?是?目外包吧,既然自己不?又不想?的?
你??功能又不??,百八十?就?有人?你搞定了.
error_reporting(0);
$conn=mysql_connect("localhost","root","root");//填写数据库连接信息
mysql_select_db("test");//填写数据库名
$valA = $_POST["valA"];
if($valA!=""){
$sql = "select * from test where a=".$valA;
$result=mysql_query($sql);
while($arr=mysql_fetch_array($result))
{
echo $arr["b"]."
";
}
}else{
?>
}
?>
谢谢~ 十分感谢
?是?目外包吧,既然自己不?又不想?的?
嘎嘎,自己用.net会,无奈知道逻辑就是变不成php的,来求教下php大神...
error_reporting(0);
$conn=mysql_connect("localhost","root","root");//填写数据库连接信息
mysql_select_db("test");//填写数据库名
$valA = $_POST["valA"];
if($valA!=""){
$sql = "select * from test where a=".$valA;
$result=mysql_query($sql);
while($arr=mysql_fetch_array($result))
{
echo $arr["b"]."
";
}
}else{
?>
}
?>
echo $arr["b"] ,b替换为我那一列的列名吗?
恩,是的,a是替换为你的条件字段名,b替换为你想获取结果的字段名

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