PHP登录错误：[重复](PHP Login error: [duplicate])

PHP登录错误：[重复](PHP Login error: [duplicate])

这个问题在这里已有答案：

mysql_fetch_array（）/ mysql_fetch_assoc（）/ mysql_fetch_row（）/ mysql_num_rows等...期望参数1为资源 31答案

我试图用php和mysql制作一个登录脚本。 一切都很好，除了我在尝试登录时遇到此错误：

警告：mysql_num_rows（）期望参数1是资源，在第10行的/home/deadmin1/public_html/login/index.php中给出布尔值

相关代码是：

<?php require("config.php"); if(isset($_POST['Submit'])){ $userName=getStr($_POST['userName']); $passWord=md5(getStr($_POST['passWord'])); $query=mysql_query("select userID from tbl_users where userName='$userName' and passWord='$passWord' "); if(mysql_num_rows($query)==1){ //login success $data=mysql_fetch_array($query,1); $_SESSION['userID']=$data['userID']; header("location:dashboard.php"); }else{ //login failed $error="Invalid Login"; } }

This question already has an answer here:

mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc… expects parameter 1 to be resource or result 32 answers

Im trying to make a login script with php and mysql. everything is fine except i get this error when trying to login:

Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in /home/deadmin1/public_html/login/index.php on line 10

the relevant code is:

<?php require("config.php"); if(isset($_POST['Submit'])){ $userName=getStr($_POST['userName']); $passWord=md5(getStr($_POST['passWord'])); $query=mysql_query("select userID from tbl_users where userName='$userName' and passWord='$passWord' "); if(mysql_num_rows($query)==1){ //login success $data=mysql_fetch_array($query,1); $_SESSION['userID']=$data['userID']; header("location:dashboard.php"); }else{ //login failed $error="Invalid Login"; } }

最满意答案

错误消息清楚地说明了查询的错误。

mysql_num_rows（）期望参数1是资源 ， boolean在...中给出

显然$query是一个布尔值而不是一个合适的mysql结果资源。 我的猜测它等于=== false 。 含义：您的查询错误：

对于SELECT，SHOW，DESCRIBE，EXPLAIN和其他返回结果集的语句，mysql_query（）在成功时返回资源，如果出错则返回FALSE。 [资源]

尝试通过检查mysql_error()和/或转储$ query var来调试查询

$query=mysql_query(....) or die (mysql_error()); or var_dump($query);

其他一些评论：

不推荐使用mysql_ *函数： faq 将此查询重写为PDO语句或mysqli_ *并不难

如果你真的想建立一个登录系统，请查看另一个算法，然后是MD5（如SHA256），salting和其他有关用户身份验证的最佳实践。

The error message clearly states what is wrong with your query.

mysql_num_rows() expects parameter 1 to be resource, boolean given in ...

Apparently $query is a boolean instead of a proper mysql result resource. My guess it equals to === false. Meaning: your query is wrong:

For SELECT, SHOW, DESCRIBE, EXPLAIN and other statements returning resultset, mysql_query() returns a resource on success, or FALSE on error. [source]

Try debugging your query by checking mysql_error() and /or dumping the $query var

$query=mysql_query(....) or die (mysql_error()); or var_dump($query);

Some other remarks:

the mysql_* functions are deprecated: faq It is not that hard to rewrite this query to a PDO statement or mysqli_*

And if you really want to make a login system, please look into an other algorithm then MD5 (like SHA256), salting and other best practices regarding user authentication.