Question

php Parse error: syntax error, unexpected T_IS_NOT_EQUAL, expecting ',' or ')' in C:\AppServ\www\0

Parse error: syntax error, unexpected T_IS_NOT_EQUAL, expecting ',' or ')' in C:\AppServ\www\003newshop\codeback.php on line 11
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<?php
include("data/database.php");
include("temp/codeback.html");
session_start();
$usernc=$_POST['usernc'];
$mb=$_POST['mb'];
$mbda=$_POST['mbda'];
$submit=$_POST['submit'];
if(isset($submit)&& $submit=="提交" ){
get_connection();
if(isset($usernc!="")){
$sql="select user from tb_reg where user='$usernc'";
$result1=mysql_query($check1,$conn);
$row=mysql_num_rows($result1);

if($row<1){
echo "<script>alert('用户名不存在!'); history.back(); </scrip>";
}
if($mb!=""){

$sql="select mbwt from tb_reg where mbwt='$mb'";
$result=mysql_query($sql,$conn);
$row2=mysql_num_rows($result2);

if($row2<1){
echo "<script>alert('密保问题不存在
!'); history.back(); </script>";
}

if($mbda!=""){
$sql="select mmda from tb_reg where mmda='$mbda'";
$result=mysql_query($sql,$conn);
$row3=mysql_num_rows($result);
if($row3<1){
echo "<script>alert('密保答案不正确!'); history.back();</script>";
}
}
}
}
close_connection();
}
?>
我没找出来

Answer 1

这好像是基本的语法错误，挨个找麻烦还不容易找到，一般编辑器都有语法错误提示 比如zend studio编辑器。 你这个错误原因是 isset 本身就是判断变量是否设置是否存在的，然后你有加了一个 !="" 把他去掉就可以了。

Answer 2

if(isset($usernc!="")){

这里哈
应该是if(isset($usernc) && $usernc != ''){
请问是否是判断存在$usernc 并且$usernc不能为空?