php 创建一个php函数来返回mysql结果
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Creating a php function to return mysql results
提问by mrpatg
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.
我试图创建一个函数，它将返回一个 mysql 查询，然后我可以循环并处理结果，但它似乎不起作用。我什至可能没有以正确的方式这样做。
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;
}
The error i recieve is
我收到的错误是
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource
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.
我尝试将上述所有内容加载到函数中，但每次只能让它返回一个结果。由于生病需要处理每个结果，我“认为”上述方式是我想要的方式，但如果有更好的方法，或者我做错了什么，请告诉我。
When i echo the function with the username, i get the following;
当我用用户名回显函数时，我得到以下信息；
Resource id #5
回答by adatapost
Remove double quotes around the link variable $result.
删除链接变量周围的双引号$result。
function GetAccounts($username){
require("dbconn.php");
$result = mysql_query("SELECT * FROM `accounts` WHERE `username` = '$username' ") or trigger_error(mysql_error());
return $result;
}
回答by Tom Haigh
Putting $resultwithin double quotes means it will be cast to a string, and is then no longer of type 'resource'. Try instead:
放在$result双引号内意味着它将被转换为字符串，然后不再是“资源”类型。试试吧：
return $result;
