警告:mysql_result():在第4行中提供的参数不是有效的MySQL结果资源(...) [英] Warning: mysql_result(): supplied argument is not a valid MySQL result resource in (...) on line 4
问题描述
非常感谢任何帮助。 :
<?php
function user_exists($ username){
$ username = sanitize( $的用户名);
return(mysql_result(mysql_query(SELECT COUNT(user_id)FROM users WHERE username = $ username),0)== 1)?真假;
}
?>
那些错误或特殊情况。 mysql_query
将返回零到n行或者出现错误。因此,返回的资源仅适用于非错误查询。这可以用来处理像下面这样的情况。
首先构建并执行查询,然后处理资源。
$ query =SELECT COUNT(user_id)FROM users WHERE username =。$ username;
$ result = mysql_query($ query);
您可以使用以下内容来确定发生错误时发生了什么:
if(!$ result)die(SELECT failed:.mysql_error());
或者这些想法来处理这个问题
if(!$ result = mysql_query($ query)){
return false; //或类似操作
}
if(mysql_num_rows($ result)!= 1){
return false;
} else {
return true;
}
Hi guys just curious to solve this annoying problem. Heres my snippet.
I've checked some other questions similar to my error but so far I cant get it solved.
Any help is much appreciated. :)
<?php
function user_exists ($username) {
$username = sanitize($username);
return (mysql_result(mysql_query("SELECT COUNT(user_id) FROM users WHERE username = $username"), 0) == 1) ? true : false;
}
?>
You should split your code in some more lines to handle those errors or special cases. mysql_query
will return zero to n rows or an error if it occurs. The returned resource will therefore only be true on non-error queries. This can be used to handle such situations like follows.
At first build and execute query, next process the resource.
$query="SELECT COUNT(user_id) FROM users WHERE username = ".$username;
$result = mysql_query($query);
u may use the following to determine what is going on in case of an error:
if(!$result) die("SELECT failed: ".mysql_error());
or these idea to handle the problem
if (!$result=mysql_query($query)) {
return false; // or similar operation
}
if (mysql_num_rows($result)!=1){
return false;
}else{
return true;
}
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