很抱歉这个愚蠢的问题,但是我看不到/不够好解决:
我有一个MySQL数据库和一个php表格,一切正常,但我不能只记录一条-可能是语法错误:
$SQL = "SELECT * FROM massfelh WHERE user = $uname AND pass = md5($pword)"; - this is working
$result = mysql_query($SQL); - this is too
$catg = $result['categ']; - help
$num_rows = mysql_num_rows($result); - this is ok too
echo $catg; - for test and I can't print it on screen
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数据库结构:id,user,pass,categ
我只需要categ(1或0),但它不会在屏幕上显示。
任何帮助将不胜感激-请说我也是愚蠢的:)
更新:
谢谢大家!!!!!对我有很大帮助。我为我的代码找到了最好的解决方案,但是所有答复都是正确的。再次感谢您,这将帮助您像我一样初学者:)再次感谢您
尝试这个..
$SQL = "SELECT * FROM massfelh WHERE user = $uname AND pass = md5($pword)";
$result = mysql_query($SQL);
$result = mysql_fetch_assoc($result);
$catg = $result['categ'];
$num_rows = mysql_num_rows($result);
echo $catg;
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