Question

我正在尝试根据数据库中的特定ID从我的数据库中获取数据。

这是我的代码。但他们没有其他工作 $selectquery,$resultsgetdata,$countprodu

<?php
$profrom = $_GET['id'];
$selectquery = "SELECT * FROM tbl_name WHERE proid = '$profrom'";
$resultsgetdata = mysql_query($selectquery);
$countprodu= mysql_num_rows($resultsgetdata);
if($countprodu>0)
{
$proidid = $row['proid'];
$proidName = $row['proName'];
$proidDescription = $row['proDescription'];
$Category   = $row['proCategory'];
$Price    = $row['Price'];
$Photo1name = $row['Photo1name'];
}

echo $proidid;
?>

Answer 1

你忘了取价值试试这个

＆＃13;

<?php
$profrom = $_GET['id'];
$selectquery = "SELECT * FROM tbl_name WHERE proid = '$profrom'";
$resultsgetdata = mysql_query($selectquery);
$countprodu= mysql_num_rows($resultsgetdata);
if($countprodu>0)
{
while($row = mysql_fetch_assoc($resultsgetdata){

$proidid = $row['proid'];
$proidName = $row['proName'];
$proidDescription = $row['proDescription'];
$Category   = $row['proCategory'];
$Price    = $row['Price'];
$Photo1name = $row['Photo1name'];
}

}


?>

＆＃13;

Answer 2

您需要使用mysql_fetch_assoc()。

<?php
$profrom = $_GET['id'];
$selectquery = "SELECT * FROM tbl_name WHERE proid = '$profrom'";
$resultsgetdata = mysql_query($selectquery);
$countprodu= mysql_num_rows($resultsgetdata);
if($countprodu>0)
{
  while($row = mysql_fetch_assoc($resultsgetdata))
  {
    $proidid = $row['proid']; 
    $proidName = $row['proName'];
    $proidDescription = $row['proDescription'];
    $Category   = $row['proCategory'];
    $Price    = $row['Price'];
    $Photo1name = $row['Photo1name'];

    echo $proidid;
  }
}
?>

Answer 3

试试这个我希望它会起作用，

<?php
$profrom = $_GET['id'];
$selectquery = "SELECT * FROM tbl_name WHERE proid = '$profrom'";
$resultsgetdata = mysql_query($selectquery);
$countprodu= mysql_num_rows($resultsgetdata);
if($countprodu>0)
{
  while($row = mysql_fetch_array($resultsgetdata))
  {
    $proidid = $row['proid']; 
    $proidName = $row['proName'];
    $proidDescription = $row['proDescription'];
    $Category   = $row['proCategory'];
    $Price    = $row['Price'];
    $Photo1name = $row['Photo1name'];

    echo $proidid;
  }
}
?>

我的mysql无法正常工作在哪里我做错了？

3 个答案: