总在if语句运行不了!请帮忙看看! 查询和插入功能!
<?php $textfield=$_request["textfield"];
$mysql_link=mysql_connect("localhost" ,"root"," '")or die("error");
mysql_select_db("bggl" ,$mysql_link);
$query="select id from xmgl where id=$textfield";
$mysql_result=mysql_query($query ,$mysql_link);
if ($mysql_result==' ')
$sql_insert="insert into xmgl values('$id', '$duty' '$move' , '$introduce', '$time')";
else
print "error";?>
问题点数：0、回复次数：3Top
1 楼flybird000（麻烦制造者）回复于 2003-06-01 18:47:58 得分 0
mysql_query 在select语句成功是返回一个资源标识符，而失败时返回false而不是空
该为if($mysql_result==false)
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2 楼wangxiangjun_sun（）回复于 2003-06-01 19:06:24 得分 0
if (!$mysql_result)Top
3 楼lapmx（e+）回复于 2003-06-01 19:23:15 得分 0
<?php $textfield=$_request["textfield"];
$mysql_link=mysql_connect("localhost" ,"root"," '")or die("error");
mysql_select_db("bggl" ,$mysql_link);
$query="select id from xmgl where id=$textfield";
//取回记录值mysql_result_num
$mysql_result_num = mysql_num_rows(mysql_query($query ,$mysql_link));
if ($mysql_result_num = 0 )
$sql_insert="insert into xmgl values('$id', '$duty' '$move' , '$introduce', '$time')";
else
print "error";?>Top
