php搜索查询数据库数据

2025-03-01 22:20:56
推荐回答(5个)
回答1:

查看一下代码:

// 获取表单提交值
$student_id = intval(trim($_POST['student_id']));
// 页面表单  可以放单独的html文件中,如果放单独的html页面中 form 的action的地址要改成下面的PHP文件名
echo '
  
  
  ';
// 当有数据提交时
if ($student_id)
{
    $con= mysql_connect("localhost","root","111") or die("连接错误");
    mysql_select_db("examination",$con);
    
    // 查询
    $sql = "SELECT * FROM tablename WHERE student_id = $student_id ";
    $res=mysql_query($sql);
    $row=mysql_fetch_array($res);
    // 输出
    echo '学号:'.$row['student_id'].'
姓名:'.$row['name'].'
性别:'.$row['gender'].'
分数:'.$row['score'];
}
?>

回答2:

$where=$_POST['student_id'];
$sql ="SELECT * FROM 表名 WHERE student _id=$where";

回答3:

$student_id = $_POST['s_id']
$sql = "select * from 表名 where student_id = ".$student_id.""

回答4:

$sql = "SELECT * FROM 表名 WHERE student _id=1000001";
$res=mysql_query($sql);
$row=mysql_fetch_array($res)

回答5:

$con= mysql_connect("localhost","root","111") or die("连接错误");
mysql_select_db("examination",$con);
$res=mysql_query($sql);
$row=mysql_fetch_array($res)
$name=$row[name];

$gender=$row[gender];
$score=$row[score];
echo $name,$gender,$score;
?>