从URL传递变量+从SQL中选择并回显它

时间:2016-10-05 04:19:27

标签: php mysql sql

标题有点难以理解,所以基本上我正在制作一个Pastebin克隆,并试图做一种viewmember.php?id = 1213用于查看贴纸的东西。但是,我根本无法弄清楚。我已经做了很多研究,在最终理解了我必须做的事情之后(或者我认为),我做了这个,不知道为什么它不起作用。

<?php
require 'connection.php';
    $getid = $_GET["id"];
    $sql = "SELECT * FROM pasteinfo WHERE id = ?";
    $stmt = $con->prepare($sql);
    $stmt->bind_param("i", $getid);
    echo $stmt;

&GT;

我可能只是愚蠢。谢谢你的帮助。

2 个答案:

答案 0 :(得分:1)

您需要运行命令来执行查询。

$sql = "SELECT field1, field2 FROM pasteinfo WHERE id = ?"; // Specify fields in query
$stmt->bind_param("i", $getid);  /* bind parameters for markers */
$stmt->execute();  /* execute query */
$stmt->bind_result($field1, $field2); /* bind result variables */
$stmt->fetch();    /* fetch value */
echo "Field 1:".$field1;
echo "Field 2:".$field2;

参考:Example #1 mysqli::prepare() example

答案 1 :(得分:0)

$resultss = mysql_query("SELECT * FROM potentiometer_sensor order by S_No desc limit 0,10");

while($rowss = mysql_fetch_array($resultss))
  {

$time1 = $rowss['2'];

$time2 = $rowss['2'];


echo "t1: ".$t1=strtotime($time1); 
echo "<br/>t2: ".$t2=strtotime($time2);    

echo "<br/>end:".$end=strtotime($time1); 
echo  "<br/>floor value:";  



var_dump (floor(((($end- $t1)/60)/60))/60);     

$Hours = floor(((($end- $t1)/60)/60)/60);    

echo   "<b>".$Hours.' </b>Hours ';

}