创建一个PHP函数以返回MySQL结果 [英] Creating a php function to return mysql results

问题描述

我正在尝试创建一个函数，该函数将返回一个mysql查询，然后我可以遍历并处理结果，但是它似乎没有用.我什至可能做不到正确的方法.

Im trying to create a function, that will return a mysql query, which i can then loop through and handle the results, but it doesnt seem to be working. I might not even be doing this the right way.

function GetAccounts($username){
require("dbconn.php");
$result = mysql_query("SELECT * FROM `accounts` WHERE `username` = '$username' ") or trigger_error(mysql_error()); 
return "$result";
}

$result = GetAccounts($username);
while($row = mysql_fetch_array($result)){ 
foreach($row AS $key => $value) { $row[$key] = stripslashes($value); } 
$theusername = $row['theusername'];
$thepassword = $row['thepassword'];
echo $theusername;
}

我收到的错误是

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource

我尝试将以上所有内容加载到函数中，但是每次只能返回一个结果.由于生病需要处理每个结果，因此我想"上述方法是我想要的方法，但是请告诉我是否有更好的方法，或者我做错了什么.

I tried loading all of the above into the function, but could only get it to return a single result each time. Since ill need to handle each result, i "think" the above way is how i want to do it, but let me know if there is a better way, or what im doing wrong.

当我用用户名回显函数时，我得到以下信息；

When i echo the function with the username, i get the following;

Resource id #5

推荐答案

删除链接变量$result周围的双引号.

Remove double quotes around the link variable $result.

function GetAccounts($username){
  require("dbconn.php");
  $result = mysql_query("SELECT * FROM `accounts` WHERE `username` = '$username' ") or trigger_error(mysql_error()); 
  return $result;
 }

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