-2

heyy guys....

i know this is probably a silly error but i really tried hard to solve this error, but can't seem to find the cause...

    $id=$_GET['uid'];
    $name=$_GET['uname'];
    $empnum=$_GET['empnum'];
    $status=$_GET['status'];
    $role=$_GET['role'];

    //--SQL query : Fetching data from main_data and how many data row exists
    $sql1=mysql_query("SELECT * FROM $tbl_name WHERE UserID='$id'");
    $count=mysql_num_rows($sql1);

    if($count>0)
    {
        echo "User ID exists";
        header("refresh:1;url=newregistration.php" );
    }
    else
    {
        //sql commands to insert the data into the database
        sql1="UPDATE $tbl_name SET UserID='$id' WHERE UserID='$id'";        //sql query defined
        sql2="UPDATE $tbl_name SET UserName='$name' WHERE UserID='$id'";
        sql3="UPDATE $tbl_name SET EmpNumber='$empnum' WHERE UserID='$id'";
        sql4="UPDATE $tbl_name SET Status='$status' WHERE UserID='$id'";
        sql5="UPDATE $tbl_name SET Role='$role' WHERE UserID='$id'";
        $result=mysql_query($sql1) or mysql_error(); 
        $result=mysql_query($sql2) or mysql_error();
        $result=mysql_query($sql3) or mysql_error();
        $result=mysql_query($sql4) or mysql_error();
        $result=mysql_query($sql5) or mysql_error();

would be grateful if someone can point out my mistake...

thank you

-- Update --

Error:

Parse error: syntax error, unexpected '=' in C:\wamp\www\ReportTrackingSystem\updateregistration.php on line 43 $sql1="UPDATE $tbl_name SET UserID='$id' WHERE UserID='$id'";
4

2 回答 2

1

你忘了把$变量 $sq1等放在前面，

于 2012-05-09T03:24:27.797 回答
1

Difficult to go off of without an exact error message, but these (sql1, sql2, etc) should be variables (flagged with $): 

$sql1="UPDATE $tbl_name SET UserID='$id' WHERE UserID='$id'";        //sql query defined
$sql2="UPDATE $tbl_name SET UserName='$name' WHERE UserID='$id'";
$sql3="UPDATE $tbl_name SET EmpNumber='$empnum' WHERE UserID='$id'";
$sql4="UPDATE $tbl_name SET Status='$status' WHERE UserID='$id'";
$sql5="UPDATE $tbl_name SET Role='$role' WHERE UserID='$id'";
于 2012-05-09T03:13:59.587 回答