Question

我需要一些帮助。我正在this页面上工作，我想知道如何显示与nick变量匹配的所有STEAMID和Nicknames。我目前使用LIKE ='％$ nick％'来查找匹配的昵称，但如果两个玩家被称为Sidewaykill，它只会显示一个，但我希望它显示全部。我并不真正关心它的外观，只要它显示所有匹配的玩家Nicknames和STEAMID。谢谢。

<?php
//Get STEAMID from steamid variable
$nick = mysql_escape_string(stripslashes($_GET["nick"]));
$steamid = $_GET["steamid"];
?>

<!DOCTYPE html>
<html><head><title>Searching for a player steamid</title>
</head>
<body>
<?php

//Connect to database

$sqluser = "----";
$sqlpass = "----";
$hostname = "localhost"; 

//connection to the database

$dbhandle = mysql_connect($hostname, $sqluser, $sqlpass) 
 or die("Unable to connect to MySQL");

//select a database to work with

$selected = mysql_select_db("versound_store",$dbhandle) 
  or die("Could not select database!");

//Attempt to find player SteamID

   $sql1 = "SELECT steamid FROM `vxp_users` WHERE nick LIKE '%$nick%'";
   $res1 = mysql_query($sql1) or trigger_error(mysql_error());
   $row1 = mysql_fetch_assoc($res1);
   $steamid2 = $row1['steamid'];

//Find latest recorded Nickname

   $sql2 = "SELECT nick FROM `vxp_users` WHERE steamid = '$steamid2'";
   $res2 = mysql_query($sql2) or trigger_error(mysql_error());
   $row2 = mysql_fetch_assoc($res2);
   $nick23 = $row2['nick'];



   if (!empty($steamid2)) {

//Write Info

    echo $nick23;
    echo '&nbsp;-&nbsp;';
    echo $steamid2;
    //echo '<br /><br />Last Recorded Steam Name:&nbsp;';
    //echo '<br /><br />Current Steam Name:&nbsp;';
    //echo $currentsteamname;
    echo '<br /><br /><a href="http://versound.org/finder/steamidtoprofile.php?steamid=';
    echo $steamid2;
    echo '" target="_blank">Link to Steam Profile</a>';


   }
   else 
   {
      echo 'The player&nbsp;';
      echo '"';
      echo $nick;
      echo '"';
      echo '&nbsp;was not found in the database. Make sure you have correctly entered their nickname.';
    }
?>
</body>
</html>

Answer 1

你需要循环：

while ($row1 = mysql_fetch_assoc($res1)) {
    echo $row1['steamid']; // or store in an array etc.
}

Answer 2

你有这条线：

$row2 = mysql_fetch_assoc($res2);

您应该用循环替换它。每当你打电话给mysql_fetch_assoc()时，它都会带来下一行 - 所以你会使用这样的东西：

while ($row = mysql_fetch_assoc($res2)) {

有关详细信息，请参阅the documentation。

警告：这些函数已从PHP 5.5弃用 - 从长远来看，使用mysqli或PDO可以让您有更好的时间。

从我的数据库中选择多个STEAMID

2 个答案: