php代码中的sql错误 [英] error on sql in php code

问题描述

亲爱的朋友们，


我正在运行代码

$ check_1 = mysql_query（" SELECT * FROM user WHERE u_name =''"。

$ _POST [''username'']。"''"）或die（mysql_error（））;


$ check_2 = mysql_num_rows（ $ check）;


但警告来了

"警告：mysql_num_rows（）：提供的参数不是有效的MySQL

结果资源在C：\Program Files \ Apache Service Foundation

\ Apache2.2 \第44行的htdocs\rajput_new\loginphp.php .."


$ _POST [''username'']在用户输入的相同代码上有表格名称

用户名..


有什么问题......

提前致谢...
$ b $原地

Dear friends,

I am running the code
$check_1 = mysql_query("SELECT * FROM user WHERE u_name = ''".
$_POST[''username'']."''") or die(mysql_error());

$check_2 = mysql_num_rows($check);

but the warning comes
" Warning: mysql_num_rows(): supplied argument is not a valid MySQL
result resource in C:\Program Files\Apache Software Foundation
\Apache2.2\htdocs\rajput_new\loginphp.php on line 44.."

$_POST[''username''] has name of form on same code the user enter
username..

What''s the problem...
Thanks in advance...
situ

推荐答案

check_1 = mysql_query（" SELECT * FROM user WHERE u_name =''"。

check_1 = mysql_query("SELECT * FROM user WHERE u_name = ''".

_POST [''username'']。"''"）或die（mysql_error（））;

_POST[''username'']."''") or die(mysql_error());

check_2 = mysql_num_rows（

check_2 = mysql_num_rows(

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